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.



Homeomorphisms vs Borel automorphisms
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...)


