Let's see, we can work this out from the albedo of the Moon, and the intensity of moonlight.

First it would be easy to arrange this by mega engineering, if we had a Moon a little larger than our Earth, because the same amount of sunlight falls on an Earth sized Moon as on Earth.

Cover the near hemisphere with motorized flat mirrors, angled to reflect almost all of its sunlight to Earth. Then, at times of full Moon, we would get as much sunlight from the Moon as we do from the midday sun, and so, the full Moon at midnight would be as bright as the sun. We need to set the mirrors to auto track the Earth, otherwise this would only work for a brief moment in its orbit.

That's a bit like the 1977 idea of Lunetta, a system of many bright mirrors in space all angled to light up a particular spot on the Earth, e.g. in the Siberian winter in Russia.

The 1977 Lunetta idea. The Russians explored a similar idea with their Znamaya 2 mirror in 1993

The same idea has been used to reflect sunlight into villages hidden in the darkness of a deep valley - the Italian town of Viganella, and the Norwegian town of Rjukan

Rjukan in Norway, hidden in a deep valley, it never gets direct sunlight in winter. An autotracking mirror now reflects sunlight from the valley rim to a bright spot in the middle of its village square.

If our Moon was a bit larger than Earth and covered in auto tracking mirrors like this, it would be as bright as the sun in our sky.

The more interesting question is, what about a natural moon not covered in mirrors. How bright can it be? Can a natural moon be as bright as the sun?
So, let's see... We can work this out from the albedo of the Moon, and the intensity of moonlight.

It's best to use the geometric albedos of planets and moons here, as that will simplify the calculations. It's the ratio of the brightness of a celestial body to the brightness of an idealised flat disk.

Geometric albedo examples (lists albedos of some of the planets and moons)

For the distance to the Moon I'll use the average distance, of 384,400 km.

LET'S TRY VENUS

Venus is one of the brighter objects in our solar system, and its quite large, nearly as large as Earth, so it seems like a good starting point. We know it can't be as bright as the sun, from the mirror tracking Moon example, but how bright can it be?

Venus

Here is the calculation, for maths geeks, indented, so that it is easy for everyone else to skip.

Our Moon has a geometric albedo of 0.12. So, actually, lunar rocks and soil are quite dark, about as dark as a typical worn asphalt road. It looks so bright in our sky because it's always lit by full sunlight. Also, especially at night, it is set against a dark sky.

Venus has a geometric albedo of 0.67, or as bright as snow, much brighter than the Moon. It's also larger.

Moon's diameter 3,476 kilometers
Venus' diameter 12,104 kilometers.

So the visual disk is increased in area by (12,104/3,476)2 and the albedo increased by 0.67/0.12. So the strength of the moonlight is increased by (12,104/3,476)2*(0.67/0.12)

(Click on the link to show the calculation in Google calculator)

So Venus would be about 68 time brighter than the moon if you replaced our Moon by Venus.

HOW MUCH DO WE NEED TO INCREASE THE BRIGHTNESS OF THE MOON BY?

Sadly sixty-eight times brighter than the Moon is not enough to make much of a difference, because the brightness of moonlight is only 0.1 lux, increasing to 0.26 lux at the tropics - see Moonlight. Sunlight ranges from 32,000 lux upwards to 100,000 Lux

So we need to increase the intensity of the moonlight 385,000 times to get the full moonlight in the tropics as bright as direct sunlight.

Light from Venus, in the place of the Moon, would have a brightness of 0.26*68 or 18 lux. That's quite bright, compare 40 lux for typical indoor lighting of a home. But nowhere near the brightness of daylight.

In all this we are not taking account of radiation in other wavelengths, such as infrared, just visible light. So we should do all our calculations using units of illuminance, such as the Lux.

HOW BIG SHOULD OUR "MOON" BE THEN?

So how big would Venus need to be to do the trick? Well as Venus was 68 times brighter than the Moon, and we need to get it 385,000 times brighter - then it needs to be (385,000/68) or about 5,700 times brighter very roughly. Or about 75 times larger in diameter. (sqrt(5,700) )

Jupiter is only 12 times wider than Venus, and what's more, it has a lower geometric albedo. So it's not going to work to use Jupiter at the distance of the Moon. But let's do the calculation anyway and see how bright it would be.

Of course by now, Earth is orbiting Jupiter rather than the other way around. Strictly speaking we are a moon of Jupiter. But for us, Jupiter looks like a Moon in our sky..

LET'S TRY JUPITER

Let's try Jupiter at the distance of the Moon.

How bright is it?

Albedo 0.52 (from Geometric albedo)
Diameter 142,984 km

So now, our calculation for the increase in the brightness of moonlight is 0.26*(142,984/3,476)2*(0.52/0.12).

(Where, as before, 0.26 is the brightness of the tropics full Moon, in lux, 3,476 is its diameter in kilometers, and 0.12 is its albedo).

or 1,906 lux

That's as bright as an overcast day, or studio lighting. So we are getting there.

To make it brighter, we can bring Jupiter closer to Earth. The area, and so the brightness, goes up rapidly, as the inverse square of the distance.

So, let's see how close Jupiter would need to be to Earth to have the same illumination as the sun.

ROCHE LIMIT

There is a limit to how close we can be to Jupiter, because if we are too close, Earth will be torn apart by tidal effects.

Any satellite that orbits within the Roche limit is torn apart by tides (probably to form a ring system).

Jupiter's closest moon is Metis (moon)

Metis, orbits with period of 7 hours minutes, faster than Jupiter's day of 9h 56m so it is being pulled inwards rather than outwards by tidal interactions and will eventually hit Jupiter. It's elongated by tidal forces as you see in this image and within the fluid Roche limit of Jupiter but must be outside its rigid Roche limit as it hasn't broken up

So how close can it be?

The Roche limit (for "fluid" satellites, which Earth counts as because the rocks are not nearly strong enough to hold it together against tidal forces of gravity) is

(D = density of parent, d = density of satellite)

If our Moon has a similar density to Earth, we can ignore the cube root as D = d, so then it becomes a simple result.

If its density is similar to Earth, then it can't orbit it closer than 2.44 times Earth's radius.

Jupiter however has a density of only 1.33 g/cm3 and Earth's is 5.51 g/cm3 so Earth can orbit rather closer than that without getting torn apart by tides. Jupiter's radius is 69,911 km.

So putting those numbers into our equation, Earth could orbit 2.44 * 69,911 * cube root (1.33/5.51) km from Jupiter

Or 106,211 km from Jupiter (distance from its centre). We'd be orbiting closer to Jupiter than its innermost moon Metis (which orbits it a distance of around 128,000 km). Earth would be seriously tidally stretched, but would survive as a planet.

So how bright is Jupiter if we orbit as close as that, only 106,211 kilometers away?

For closely orbiting planets, it's not accurate enough to just estimate the increase in brightness using the inverse square of the distance. The near side of the planet will be a lot closer than that and will fill much more of the sky.

We need to use the area subtended by a sphere to do this accurately. I'll indent this calculation so you can skip it easily if not interested in the details:

The exact formula is that the solid angle is 2 * PI * (1-sqrt(r2 - R2)/r) for r > R where r is the distance from the center of the planet and R is its radius. See Solid angles subtended by rotary solids at a point.

So we need to use the ratios of the solid angles at r = 106,210 km and at r = 384,400 km (distance of the Moon), where R (Jupiter's radius) = 69,911 km. The 2*PI will cancel so can ignore it when we take the ratio.

Using the solid angles formula, then its brightness increases from 1906 lux to

1906 * (1-sqrt( 106,2102 - 69,9112) / 106,210)/( (1-sqrt(384,4002 - 69,9112)/384,400))

Or, 28,250 lux

That is for Jupiter with its usual geometric albedo of 0.52. If it had the same surface brightness as Venus, then its brightness would be (0.67/0.52)*28,250 or about 36,400 lux.

This is well into the lower range for the brightness of full sunlight, of 32,000 to 100,000 lux.

What about a larger planet?

LARGEST EXOPLANET - LET'S TRY WASP-17b

What if it is less dense than Jupiter Here is the largest exoplanet discovered

WASP-17b

First, as for Jupiter, we need to work out how closely Earth can orbit it safely.

The figures in that report are given relative to Jupiter. Its radius is 1.991 times the radius of Jupiter, which makes it 1.991*69,911 km, so roughly 140,000 km. So, its diameter is around 280,000 km.

Its density is 0.486 times the density of Jupiter so from Jupiter's density, that's 0.486*1.33. or about 0.65.

Earth's density d is 5.51 g/cm3

So its Roche limit for Earth is 2.44 * radius * cube root (ratio of densities)
= 2.44 * 140,000 * cube root (0.65/5.51)

= 167,500 km.

So we could orbit WASP-17b at less than half the distance to the Moon and not be torn apart. Indeed we'd be skimming just 27,500 km above its surface, at the Roche limit.

First, let's just leave it at the Moon's distance for simplicity.

The only thing is I don't know how bright it is, its geometrical albedo - we could probably calculate that from its distance and magnitude. But meanwhile let's just suppose it is the same as Venus.

Then we get its brightness at the distance of the Moon

as 0.26*(280,000/3,476)2*(0.67/0.12)

(same formula as before, 0.26 is the brightness of the tropics full Moon in lux, 3,476 is its diameter in km, and 0.12 is its albedo - here 280,000 is the approximate diameter of WASP-17b).

Or about 9,419 Lux. So that's about the same brightness as full daylight but not direct sunlight. Its brighter than an overcast day. Radiometry and photometry in astronomy

So - that's actually not as bright as a close orbit around Jupiter. On the other hand it is probably safer to orbit at some distance from a giant planet, less likely to be fried by fierce electromagnetic radiation perhaps.

We can estimate its visual diameter from the visual diameter of the Moon of 0.5 degrees, as (278,385 / 3,476)*0.5 or about 40 degrees.

So it would fill a large part of the sky.

Coincidentally its diameter of 280,000 km is roughly the same as the diameter of Saturn's ring system 270,000 km.

The last image in this sequence gives an idea of what it would be like to replace our Moon by the largest known exoplanet. Imagine a planet as wide in diameter as Saturn is in this video, and you've got it.

If we bring it closer, to the Roche limit for Earth - as close as we can get it without the Earth breaking up, again we need to use solid angles as for Jupiter, and as before, lets indent the calculation:

Solid angle = 2 * PI * (1-sqrt(r2 - R2)/r) for r > R where r is the distance from the center of the planet and R is is radius (by the formula for the solid angle of a sphere subtended at a point).

So we need to use the ratios of the solid angles at r = 167,500 km and at r = 384,400 km (distance to the Moon), where R (WASP-15b's radius) = 140,000 km. Using the solid angles formula, then its brightness increases from 9,419 lux to

9,419* (1-sqrt( 167,5002 - 140,000 2) / 167,500)/( (1-sqrt(384,4002 - 140,000 2)/384,400))

Or, about 61,850 lux. A fair bit brighter than Jupiter and close to the midday tropical sun at 100,000 lux.

How large would it look in the sky?

Its solid angle is 2*PI*(1-sqrt(167,5002 - 140,0002)/167,500), or about 2.834 radians. Converting that to the angle subtended (details of how they are related here), that's

2*acos(1-2.8337/(2*PI)) = 1.979 radians,

or in degrees, about 113 degrees.

So it would pretty much fill the entire sky, as expected since we are skimming just 27,500 km above its surface.

Only thing is, that the Earth will cast a significant shadow on it, at every "full Moon" probably, which is hard to calculate.

It will probably also have significant limb darkening as seen from Earth.

And Earth would be tidally locked to it, so only one hemisphere would see this bright Moon in our sky.

Anyway - as a rough estimate it seems we have pretty much achieved our goal.

We can also work out the orbital period of Earth around the planet.

Its mass, at 0.486 Jupiters is about 0.992e+27 kg (where e+27 means its multiplied by 1027, I show it in that notation so you can copy / paste it into the online calculator),

so entering this number for the mass of the "sun" into this calculator, and using the Semimajor axis as semimajor axis 165500 km, and mass of the "planet" as one Earth.

We get a period of 14.6601 hours

TYPICAL DAY ON A TIDALLY LOCKED EARTH CLOSELY ORBITING A SUPER JUPITER IN ITS SKY

If you lived on the Super Jupiter facing side, then it is fixed in position in the middle of the sky above your head, day and night, pretty much filling your sky, and you'd get this sequence, starting from midday:

1. Midday. Super Jupiter is not visible in the midday sky because it is in "new moon phase". But around midday every day, you'd get a long eclipse with the sun hidden behind Super Jupiter. So that would be the only time of darkness in the day. That's the only time when you'd be able to see the stars, though they'd be partly blocked by a big black patch in the centre of your sky in the direction of the Super Jupiter. If you are in that close orbit around Wasp-15b, the black patch would nearly fill the sky and you'd only see stars within 34 degrees of the horizon.

2, As the sun sets, Super Jupiter catches the sunlight and gets brighter. When the sun dips below the horizon Jupiter is already at first quarter, so it is half lit, and you already get full daylight from it, nearly as bright as a midday tropical sun.

3, At midnight you get "full Super Jupiter" (every "night") - but as midnight approaches, the shadow of the Earth would move across Jupiter and at midnight exactly, this shadow would be directly overhead. So, you'd get a dip in brightness at that moment. Since it is so close, this dark patch would be a large one. Though the edges of the Super Jupiter itself would still be bright around the horizon, as our shadow only shades a spot in the center.

4. Then after that the Super Jupiter wanes and the sun rises.
That whole cycle repeats every orbital period, every fourteen hours forty minutes approximately in the case of Wasp-15b, with darkness only at midday during the long solar eclipse you get every day.

On the other side of the Earth facing away from Super Jupiter, you would get no light from it at all. You just get a fourteen hour forty minutes day with the sun rising and setting normally.

Let's look briefly at other ways of doing it now. If our "Moon" had rings, it could be even brighter.

A "MOON" WITH RINGS

Those would have no effect on the Earth's Roche limit as they are of negligible mass compared to the planet (or "Moon") itself.

Saturn varies in brightness by about 0.9 magnitudes for different oppositions (times when it is closest to Earth) depending on whether its rings are open or closed. So the Saturn rings, when open, more than double the brightness of the planet. It's about 2.29 times brighter when the rings are fully open (using the formula for magnitudes, as 2.510.9 = 2.29 ). At those times, Saturn is tilted by 27 degrees.

If it was like this giant ring system, 200 times larger than Saturn it would have a larger effect:

This is an artist's impression of the giant ring system Super-Saturn J1407b   -   "Super Saturn" Has an Enormous Ring System and Maybe Even Exomoons - 200 times larger than Saturn's ring system,  120 million kilometers Super-Saturn J1407b Hosts Massive Ring System, Astronomers Say

However this is a very young system in process of forming satellites, its sun is just 16 millions years old. Faraway planet J1407b is lord of the rings. So the rings will probably dissipate as the moons form.

Rings like this would surely have a huge effect if they filled our sky. The only problem is though, that our orbit would most likely be in the ring plane.

It would help if we were on a ring crossing orbit, and as perpendicular to the rings as possible. Saturn does have some moons in highly inclined orbits.

But they tend to be further away.

As a large planet orbiting close to the ringed planet, we would probably orbit in the same plane as the rings. So we wouldn't get much sunlight reflected.

AREA OF SKY TO COVER TO GET SAME BRIGHTNESS AS FULL SUNLIGHT

If we assume the rings are as bright as they possibly could be pretty much - as bright as Venus, and we are in a polar orbit over the poles, and the sun is directly behind us shining on the rings, then we can do the calculation, this would be the brightest they get, twice a year.

For 32,000 Lux then we need 32,000/8,338 or 3.8 times as much visual area of the sky as WASP-17b. So diameter is sqrt(100,000/8,338) *278,385  = 542,672 kilometers, and it would span  (542,672 /3,476)*0.5 or 78 degrees of the sky.

Or for the 100,000 Lux - bright sunlight with sun overhead in tropics - then its sqrt(100,000/8,338) *278,385 or 964,084. Visual diameter when we are opposite the ring system:  (964,084 /3,476)*0.5 or about 139 degrees

So (if I've got these calculations right) an almost opaque ring system a million kilometers in diameter like J1407b would give us similar illumination to direct sunlight, and meanwhile it would pretty much fill the sky when overhead.

But a ring system as large as that would probably be a young system like J1407b and would soon form satellites and dissipate over geological timescales.

GRAVITATIONALLY LOCKED BINARY PLANET

(Image NASA) Are Habitable Binary Planets Possible? : DNews -  'Double Earths' Could Be Fun Exoplanets To Hunt For -- If They Exist
see: Can binary terrestrial planets exist?

Explored fictionally in Robert Forward's "Rocheworld"

Rocheworld

If the two planets are gravitationally locked to each other, in the same way as Pluto and Charon, then the tidal effects no longer matter.

They are just permanently distorted. You can even have an Earth sized planet almost touching our Earth - even with a shared atmosphere and ocean in principle

We don't know any contact binary planets, but many contact binary stars are known, and including "overcontact binaries" that are so close together that their atmospheres overlap - the first discovered W Ursae Majoris.

from: Chapter 19-7

see Contact Binaries

So the "Roche world" scenario certainly seems physically possible.

So in that case, well forget about all those limitations. It could be as big as Jupiter and only 69,911 kms away. If we and Jupiter are tidally locked, then no problem.

In a contact or over contact binary, then if you lived close to the contact point, the other planet would fill your entire sky. On the other hand at midday, as we saw before, indeed for much of the day, every day, our "Moon" would eclipse the sun.

So, the calculations get more complicated to figure it out exactly - how bright it would be at all times of the day at various distances from the contact point.

For instance if you lived right at the contact point, you'd only see the sun at sunrise and sunset, and the rest of the time it would be hidden behind both planets, and whichever planet it wasn't illuminating would be in eclipse.

In that situation, your main illumination would be at "night" when the sun would illuminate the opposite planet around your horizon.

So someone would need to do a proper detailed analysis of this. But it seems likely that with a contact binary with the other planet filling your sky, there would be places close to the contact point and times of day when it was even brighter than for our Wasp-15b calculation.

We can work out the orbital periods easily though.

Io orbits Jupiter at a distance of 420,000 km in  42.5 hours .

By Kepler's third law,

(42.5)^2/(420,000 )^3 = P^2/(69,911)^3

so if Earth was orbiting Jupiter so close that it touches it, it's period would be

P = sqrt((69,911)^3*(42.5)^2/(420,000 )^3)
= 2.89 hours

So if Earth was a contact binary with Jupiter, it's orbital period and Jupiter's rotation period would both be 2.89 hours.

Let's now try a Roche world scenario using Earth with WASP-17b.

It has half the mass of Jupiter and twice the radius. So we'd orbit it at twice that distance, so that gives us sqrt((2*69,911)^3*(42.5)^2/(420,000 )^3) or 8.1635 hours for the period. But then half the mass, makes it slower by sqrt(1/0.486) (period squared is inversely proportional to the mass of the star) so that's 11.71 hours (see Gravity Applications (for the derivation of formula for orbital period))

So, now we have a much longer orbital period of nearly 12 hours instead of under 3 hours.

So if we were orbiting WASP-17b, with us and WASP-17b both tidally locked, and WASP-17b had a day of 11.71 hours - a little under 12 hours - then that would be stable, and we wouldn't be ripped apart.

Then the side of Earth towards it would get cycles of brightness all the way to full sunlight equivalent and then down to none every twelve hours. And it would be a planet with a constantly changing phase fixed in position in the sky - and the sun would set every six hours approx.

A PLANET THAT LIGHT'S UP ITS OWN SKY

It's possible in principle to get a planet that takes up non convex shapes, if it spins fast enough. If so then you might get your planet appearing in its own sky.

For instance if a planet spins fast enough, it can take many different shapes theoretically, including a donut shaped planet.

You can also get donut shaped solutions for end state of a collapsing gas cloud (so also in principle young stars could be toriodal): 1992ApJ...401..618N Page 618

So, though it may seem unlikely, still, doesn't seem you can rule out a donut shaped planet. (Or for that matter maybe such a planet created artificially by some ETI with mega technology)

If you lived near the hole in the donut in a donut shaped Earth you'd get a lot of reflected light from the planet itself

And it could have moons also adding to the brightness:

Just mentioning it for completeness. As we don't know of any planets shaped like this and it would probably need very rare and unusual conditions to create one, perhaps we can leave it for now.

Anders Sandberg's fun article explores many aspects of life on a donut shaped planet - and the orbits of satellites and moons around such a planet.

What would the Earth be like if it was the shape of a donut?

See also: Robert Walker's answer to How and why are planets spherical? What makes them round?

HABITABLE ZONE

So anyway that's an interesting thought. What about a moon orbiting a big gas giant like WASP-17b just outside the habitable zone. It might get a fair amount of light from the gas giant - not quite as much as comes from its sun but getting on for it. I wonder if that extends the habitable zone outwards for moons of gas giants? I think it would, slightly,

COMMENT IF YOU FIND ANY MISTAKES IN THIS "BACK OF THE ENVELOPE" CALCULATION

This is just a first rough calculation, if you spot any mistakes in this calculation do say!

I found one immediately after writing, first attempt I thought WASP-17b  would be as bright as direct sunlight, but turns out it is brighter than an overcast day but not as bright as direct sunlight if I got this right.

Then after another calculation, I found it can be moved closer to the Earth than I'd thought without disrupting it,  making it as bright as the sun after all.

I wrote this up for my science blog on Science20 here:

Can A Planet's Moon Be As Bright As Its Sun?

And you can also get this and many more of my answers now as a kindle book:

Simple Questions - Surprising Answers - In Astronomy