Let's see how big the mirror would have to be at Mars to be visible with a telescope a (serious) amateur might have. The distance between Earth and Mars varies from around 60 million km to 400 million km. So we have to take into account the diminution with distance compared to the lunar distance of 380,000 km. However, the distance of Mars from the Sun is also larger than the Earth's so the reflected image would be dimmer because of that as well.
The furthest distance of Mars from the Sun is about 250 million km. This is about 1.6 times as far as the Earth distance so the mirror reflection would be dimmed by a factor of 1.6^2 = 2.56 to begin with. At the closest Earth-Mars distance of 60 million km, this is 158 times
further than the Earth-Moon distance so the reflection seen at the Earth would be dimmer by an additional factor of 158^2, about 25,000 times dimmer. At the furthest Earth-Mars distance of 400 million km, this is 1053 times further than the Moon, so it would be dimmed by a
factor of 1,100,000 times.
So the total diminution at Mars' closest approach would be (1/2.56)x (1/25,000) = 1/64,000. And at furthest distance, it would be (1/2.56)x(1/1,100,000) = 1/2,816,000. The closest distance dimming of 1/64,000 amounts to a change of apparent magnitude of 12, so to +15 for the
same 4 m sized mirror. The furthest distance dimming of 1/2,816,000 amounts to a change of apparent magnitude of 16 so to +19.
This article gives the apparent magnitude an 8-inch telescope might detect as +14 under good seeing conditions:

A practical guide to buying telescopes.
By Jeff Kanipe
Special to SPACE.com
posted: 04:05 pm ET
19 June 2000
http://www.space.com/scienceastronomy/astronomy/telescope_II.html

So at Mars closest approach you would need a larger reflecting surface to amount to a change in apparent magnitude of +1, so one of 2.51 times greater collecting area than the 4 meter one, so to 6.4 meters across. However, for the furthest distance magnitude of +19 you would need a larger reflecting surface than the 4 meter one to amount to a change in apparent magnitude of +5 in order to be visible by the 8-inch scope. So it would need to be larger by a factor of 100 in collecting area, so to 40 meters across.
Certainly a large reflective surface. But the Echo 2 satellite used a microwave reflective balloon surface of 41 meter diameter in the 1960's so it is technically feasible:

Echo satellite.
http://en.wikipedia.org/wiki/Echo_satellites

The image size as before would be calculated from the image and object distances, and the Sun's diameter. It would range from about 1/4 the Sun's size to about 1-1/2 times the Sun's size. This again raises the question: would all observers within the large observing region really see the reflecting surface as having those apparent magnitudes calculated?

Bob Clark