First: I don't know if I'm using the terminology in the right way, since I translated this.
In my textbook, a congruence is a transformation that keeps all distances equal, and it is proved that this is either a translation, rotation or reflection by a line through the origin.
Prove that there are $48$ congruences of $\mathbb{R}^3$ that map a given cube into itself. Do these form a group? An abelian group?
Well, I tried to solve as follows: there are four diagonals for the cube, which may be mirrored or might stay equal, giving $4!=24$ different congruences. Then, just like in a dihedral group, we can, for all of these congruences, invert the cube, so that other points in $\mathbb{R}^3$ are mirrored by the plane $z=0$, where $(x,y,z)$ are points in $\mathbb{R}^3$. Thus we find $24\cdot2=48$ congruences.
It is easy to see that this is a group: we have the identity image and can easily invert all rotations and reflections. Associativity also follows easily. Commutativity is not true, because this is not even the case for a dihedral group in $\mathbb{R}^2$.
I think this answer is allright, but then the follow up question:
How many congruences of $\mathbb{R}^3$ that map a given regular polyhedron with $20$ sides into itself?
I really have no idea where to start. Wikipedia tells me that I'm looking for a regular icosahedron. As I interpret this, there should be $120$ such congruences. But I don't see how I should derive this and how I would know, without the use of internet, what a regular polyhedron with $20$ sides would look like. Any help is much appreciated!
N.B.: this question appears in the first chapter of a group theory syllabus/textbook, including definitions, examples and basic proofs about groups, so the answer should be elementary.