Form of isomorphisms between subgroups of GLn Suppose that I've been given two subgroups of the general linear group (over the real numbers), which are isomorphic. From this information alone, can I deduce the form of the isomorphism? I suspect that there should be an invertible matrix $S$ such that each matrix $A$ gets mapped to $S^{-1}AS$. Is that correct? How could I prove it?
Since the original question has been answered, I like to rephrase it a bit: Suppose the subgroups are in fact Lie groups. Under which conditions on the subgroups is my conclusion valid?
 A: I'm afraid this is false.
A mapping of the form $A\mapsto S^{-1}AS$ will give the identity mapping on the sets of determinants of the elements of the group. This alone gives us the following two counterexamples:
Let $G$ be the subgroup of $GL_2(\mathbf{R})$ generated by the matrix $\pmatrix{2&0\cr0&1\cr}$, and let $H$ be the subgroup generated by the matrix $\pmatrix{1&0\cr1&1\cr}$. Both are infinite cyclic groups, so they are isomorphic. However, the determinants of all the matrices in $H$ are equal to one, but the determinants of the elements of $G$ are of the form $2^n, n\in\mathbf{Z}$.
Things won't improve, if we restrict ourselves to finite groups. Either one of the matrices
$\pmatrix{-1&0\cr0&-1\cr}$ or $\pmatrix{-1&0\cr0&1\cr}$ will generate a cyclic group of order 2. But $-1$ is a determinant of only the latter group.
There are counterexamples even of such finite groups that have the same multiset of determinants. If a finite group has two non-isomorphic faithful real representations of the same dimension, then with a little bit of luck the multisets of determinants (consisting obviously of $\pm1$:s only) are equal, and we are done. The first example that comes to mind is the cyclic group of order 4. Either
$$
\pmatrix{0&1&0&0\cr-1&0&0&0\cr0&0&1&0\cr0&0&0&1\cr}\quad \text{or}\quad
\pmatrix{0&1&0&0\cr-1&0&0&0\cr0&0&0&1\cr0&0&-1&0\cr} 
$$
will generate such a group, so they are isomorphic. This time the determinants are all ones. However, the multisets of traces differ. Recall that $tr(S^{-1}AS)=tr(A)$.
Here the traces of the elements of the group generated by the first matrix are $4,2,0,2$ and those of the group generated by the second matrix are $4,0,-4,0$, so no isomorphism of the prescribed type can exist.

Only if the groups have the property that even the traces of the matrices match up under the isomorphism (elementwise, i.e. the matrices paired up by the isomorphism have the same trace), then representation theory of finite groups promises the kind of isomorphism that you are looking for. Edit: the presence of outer automorphisms may kill this argument. Striking it for now.
