Homeomorphisms vs Borel automorphisms Let $\mathrm{Homeo}(M)$ and $\mathrm{Borel}(M)$ be the groups of homeomorphic and Borel automorphisms of a space $M$, respectively.
Question: Are $\mathrm{Homeo}(M)$ and $\mathrm{Borel}(M)$ isomorphic as abstract groups for any reasonable non-discrete space $M$? Say, $M=\mathbb{R}$ or $M=\mathbb{S}^1$. 
Obviously $\mathrm{Homeo}(M)\subsetneq\mathrm{Borel}(M)$ if $M$ is non-discrete, but with infinite cardinality this does not prohibit an algebraic isomorphism. The literature on $\mathrm{Homeo}(\mathbb{R})$ and $\mathrm{Borel}(\mathbb{R})$ that I have seen so far has not been explicit enough for me to figure out the answer in this case.
Thank you.
 A: 
For every uncountable Polish space $M$ (hence every non-discrete metrizable manifold with countably many components), the groups $\mathrm{Homeo}(M)$ and $\mathrm{Borel}(M)$ are non-isomorphic. Furthermore, $\mathrm{Borel}(M)$ is not isomorphic to any subgroup of $\mathrm{Homeo}(M)$.

Indeed, $\mathrm{Homeo}(M)$ is a separable metrizable group, and I claim that the subgroup $\mathfrak{S}_{\mathrm{fin}}(M)$ of finitely supported permutations of $M$ (which is a subgroup of $\mathrm{Borel}(M)$) does not embed into any separable metrizable topological group $G$. Indeed, let $(M_i)_{i\in I}$ be pairwise disjoint finite subsets of $M$, each of cardinal $\ge 3$, with $I$ uncountable, and let $\Gamma_i$ be a non-abelian subgroup of permutations of $M_i$, extended to the identity outside $M_i$. 
By contradiction, let $f$ be an embedding of $\mathfrak{S}_{\mathrm{fin}}(M)$ into $G$. Since every subset of a separable metrizable space is separable, there exists a countable subset $J$ of $I$ such that $f\Big(\bigcup_{j\in J}f(M_j)\Big)$ is dense in $f\Big(\bigcup_{i\in I}f(M_i)\Big)$. Choose $i\in I-J$. Then $[M_i,M_j]=\{1\}$ for all $j\in J$, so $[f(M_i),f(M_j)]=\{1\}$ . By density, we deduce $[f(M_i),f(M_i)]=\{1\}$. But the latter equals $f([M_i,M_i])$, which is not trivial since $M_i$ is non-abelian and $f$ is injective. Contradiction.

For countable Polish spaces, one can give a full answer too: basically Goldstern's example is the only one along with discrete ones.

Let $M$ be a countable Polish space. Then $\mathrm{Homeo}(M)$ is non-isomorphic to $\mathrm{Borel}(M)$ (which equals $\mathrm{S}(M)$, the whole permutation group), with the only exceptions when $X$ is finite, infinite discrete or the 1-point compactification of an infinite countable set. Namely, beyond these exceptions, $\mathrm{Homeo}(M)$ is infinite and has at least 5 normal subgroups.

For $M$ infinite, by the Onofri-Schreier-Ulam theorem, $\mathrm{S}(M)$ has exactly 4 normal subgroups (trivial, whole, finite support, even finite support). In $\mathrm{Homeo}(M)$ we already have such a chain, where the intermediate subgroups are the group $\mathfrak{S}_{\mathrm{fin}}(M_{\mathrm{iso}})$ of finitely supported permutations of the subset $M_{\mathrm{iso}}$ of isolated points of $M$ ($M_{\mathrm{iso}}$ is dense in $M$, hence infinite) and its index 2 subgroup. Hence, if $\mathrm{Homeo}(M)$ is homeomorphic to $\mathrm{S}(M)$, then every normal subgroup is one of these.
Let $M_{\mathrm{acc}}$ be the set of accumulation points.
If $M_{\mathrm{acc}}$ is empty then $M$ is discrete and this case is clear. Next we assume $M_{\mathrm{acc}}$ non-empty.
Let $H_M$ be the group of those self-homeomorphisms of $M$ that are identity on $M_{\mathrm{acc}}$, so $\mathfrak{S}_{\mathrm{fin}}(M_{\mathrm{iso}})\subset H_M$. Hence we either have (a) $H_M=\mathrm{Homeo}(M)$ or (b) $H_M=\mathfrak{S}_{\mathrm{fin}}(M_{\mathrm{iso}})$. Also denote $G_x$ as the subgroup of $\mathrm{Homeo}(M)$ consisting of those $g$ that are identity at the neighborhood of $x\in M$; if $x$ is $\mathrm{Homeo}(M)$-invariant then $G_x$ is a normal subgroup.
Lemma: for every topological metrizable space and point $x_0$ which is isolated among accumulation point and neighborhood $V$ of $x_0$, there exists a self-homeomorphism fixing $x_0$, not with finite support, and identity outside $V$.
Indeed, one can suppose that $V\smallsetminus\{x_0\}$ consists of isolated points of $X$; choose an injective sequence $(y_n)$ tending to $x_0$. Then permuting $y_{2n-1}$ and $y_{2n}$ and fixing all the remainder yields the desired permutation.$\Box$
Choose $x_0$ an isolated point in $M_{\mathrm{acc}}$. Applying the lemma to $x_0$, we see that  $H_M$ is not reduced to $\mathfrak{S}_{\mathrm{fin}}(M_{\mathrm{iso}})$, which excludes (b). So assume (a): $\mathrm{Homeo}(M)$ acts trivially on $M_{\mathrm{acc}}$. 
In particular, $G_{x_0}$ is a normal subgroup; by the lemma it is a proper subgroup of $\mathrm{Homeo}(M)$. If we assume that $X$ is not the 1-point compactification of a discrete set, we either (a') have another point in $M_{\mathrm{acc}}$, and hence another isolated point in $M_{\mathrm{acc}}$ or (b') there exists an open infinite discrete subset in $M$. In both cases, we deduce (using the lemma in Case (a')) that $G_{x_0}$ is not reduced to $\mathfrak{S}_{\mathrm{fin}}(M_{\mathrm{iso}})$.
A: Let $\bar X = X\cup \{\infty\}$ be the 1-point compactification of a discrete space $X$.  Then the autohomeomorphisms of $\bar X$ are all permutations $p$ of $X$ (extended by $p(\infty)=\infty$), whereas the Borel automorphisms are all permutations of $\bar X$ (since every subset of $\bar X$ is Borel, even open$\cup$closed).  Algebraically, these two groups are isomorphic.  
(But is this space "reasonable"?  Felix Hausdorff might disagree...)
