Suppose that we have two closed n-manifold $M$ and $N$ such that the topological group of homeomorphisms $Homeo(M)$ is homotopy equivalent to $Homeo(N)$ (maybe as topological groups if needed), can we deduce that $M$ is homotopy equivalent (or even homeomorphic ) to $N$ ? Is there an easy counterexample ?
It is a result of Whittaker (1963), and Rubin (1990, in greater generality with a better proof) that $Homeo(M)$ (viewed as a group) determines $M$ up to homeomorphism (see this question). I doubt that the homotopy type is enough: Gabai had shown that for closed hyperbolic 3-manifolds, the inclusion of $Isom(M)$ into $Diff(M)$ is a homotopy equivalence. $Isom(M)$ is a finite group, so its homotopy type is given by its order. I don't think it is hard to find two hyperbolic manifolds with the same cardinality of isometry group (in fact, I assume that order is usually equal to $1.$)