Why is the electric field around a hollow spherical conductor homogeneous, even with an off-center charge inside? I have a lot of records in EM, and I know all about charge induction and Gauss's theorem for systems of conductors, nonetheless I still have a problem that I cannot face without feeling uncomfortable. 
Suppose to have a hollow conducting sphere, with a pointlike charge $q$ inside, placed at a point not at the centre of the sphere. That induces an asymmetrical (but axisymmetrical) charge distribution on the inner surface of the hollow sphere; but also, a perfectly homogeneous charge distribution on the external surface. Why is this?
This is something I can understand could happen, but I miss some actual proof that it must happen. It must reside in something related to the particular symmetry of the sphere, but for me it is not enough to say that "this happens due to the spherical symmetry". Is there something that clearly forces things to happen like this?
 A: Since there must be no field inside the metal of the sphere, the charges on the inner side arrange themselves to exactly cancel the field of the point charge q.
The charges on the outside thus don't feel any field except themselves. They arrange uniformly around the sphere.
In the end, the reason why the field is independent on the exact position of the innermost charge is that it is shielded by the metal cage.
A: The metal of the conductor 'shields' the outer surface charges from the inner ones, because no macroscopic electrostatic field can exist inside the metal of the conductor.As such, the outer charges have no information about the presence of inner charges. So, the charges must exist in the form so as to make the conductor surface an equipotential(since this is the lowest energy configuration). For a sphere, this is simply uniform due to homegeneity of space. It may not be uniform for another random shape, but it always MUST be an equipotential.
A: From what I figure you're describing a shell of metal sphere enclosing some charge. It doesn't matter if you place the charge at the centre or not. The external surface charge distribution will always be uniform. If you place charge Q inside, it'll induce -Q on the inner surface and the inner surface charge distribution will be non-uniform depending on the location of the free charge. Now if you evaluate $\oint \vec E . d\vec l = 0$ on a curve passing through the shell's 'meat' and inside of of the sphere you must get that the shell itself has no field inside. Only charge has been lost from the shell to the inner surface. The total amount of charge lost is then pulled away from the external surface and if the external surface would have non-uniform charge distribution then it means there would be a tangential current flowing from the patch having higher charge than it's surroundings until the surroundings and the patch are at equipotential and thus you can't calculate where the charge inside is located. 
A: This is really a symmetry argument but I think answers your question?
The spherical conducting shell is an equipotential.
A charge outside that shell can only feel the effect of the charges on the outside surface of the shell there being no electric field inside the conducting shell.
If a charge, which starts at infinity, is moved to the surface of the shell the work done must be independent of the path taken.
If the surface charge density was not uniform over the whole sphere the work done in moving the charge to the sphere would not be independent of path.  More work would be done moving the charge towards a region where the surface charge density was larger.
