$M_1,M_2$ not homeomorphic but $G(M_1),G(M_2)$ isomorphic. 
Let $M$ be a metric space, and let $G(M)$ denote the set of homeomorphisms of $M$ onto $M$. Then $G(M)$ is a group under composition. Show that if $M_1,M_2$ are homeomorphic metric spaces, then $G(M_1)$ is isomorphic to $G(M_2)$. Give an example to show that the converse does not hold.

I settled the first part routinely. For the example, I tried to think about simple examples, like when $M_1, M_2$ are finite and have the same size. In this case, the homeomorphisms are just permutations. However, $M_1,M_2$ are homeomorphic, since any bijection from $M_1$ to $M_2$ is continuous. For other choices of $M_1,M_2$, the groups $G(M_1),G(M_2)$ seem much harder to identify.
 A: A rigid space is one whose only homeomorphism onto itself is the identity. The case $\kappa=\omega$ of Theorem $2$ of this old paper of mine immediately implies that $[0,1]$ has $2^\omega=\mathfrak{c}$ pairwise non-homeomorphic rigid subspaces, which of course all have trivial autohomeomorphism group. (This is admittedly using a piledriver to swat a gnat.)
Added: Here’s a more accessible example. Let $M_1=\left\{\frac1n:n\in\Bbb Z^+\right\}$ with the usual metric, and let $M_2=M_1\cup\{0\}$, also with the usual metric. The autohomeomorphisms of $M_1$ are clearly the permutations of $M_1$. Now suppose that $h:M_2\to M_2$ is an autohomeomorphism; clearly $h(0)$ must be $0$, since $0$ is the only non-isolated point of $M_2$. Thus, $h\upharpoonright M_1$ is an autohomeomorphism of $M_1$. To finish the example, just show that if $f:M_1\to M_1$ is any bijection, the map
$$h:M_2\to M_2:x\mapsto\begin{cases}
f(x),&\text{if }x\in M_1\\
0,&\text{if }x=0
\end{cases}$$
is an autohomeomorphism of $M_2$; it follows immediately that $G(M_1)$ and $G(M_2)$ are both the permutation group on a countably infinite set.
A: Take  $M_1=[-1,1]$ and $M_2=(-1,1)$.  The restriction map from $G[-1,1]$ to $G(-1,1)$, $f\mapsto f|_{(-1,1)}$, is an isomorphism.
A: Hint: Let $M$ be the disjoint union of the non-homeomorphic connected spaces $M_1$ and $M_2$. Can you write $G(M)$ in terms of $G(M_1)$ and $G(M_2)$?
EDIT: Okay, apparently this was too obscure, and other people have posted examples anyway. Let $M_1$ be your favorite connected metric space containing at least two points and $M_2$ be the disjoint union of that space with a point. Then clearly $M_1$ and $M_2$ are not homeomorphic, but $G(M_1) \simeq G(M_2)$.
