Suppose $X$ is a CW-complex. The monoid of homotopy self-equivalences $M = hAut(X)$ is the subspace of $Map(X,X)$ consisting of those maps with a homotopy inverse. It is a union of path components. It obviously acts on $X$, and the homotopy type only depends on the homotopy type of $X$.

It is known that we can find maps $G \leftarrow M' \to M$ of topological monoids, all homotopy equivalences, with $G$ a topological group.

It is also known that we can use this to find a $G$-space $Y$ and maps of $M'$-spaces $Y \leftarrow X' \to X$ which are all homotopy equivalence. In other words, this rigidifies the action of $hAut(X)$ to an honest action of a topological group.

However, even in this situation we have a composite map $G \to Aut(Y) \to hAut(Y)$ that we know is a homotopy equivalence, but it is unlikely to be the case that $Aut(Y)$ is homotopy equivalent to $hAut(Y)$.

Can we rigidify this and find an $Y$ whose automorphism group is equivalent to its homotopy automorphism group? Or does there exist a space for which the map $Aut(Y) \to hAut(Y)$ is *never* a homotopy equivalence for any space homotopy equivalent to $X$?

simplehomotopy type of $X$. If $X$ is simply connected, then $Aut(X\times Q)\to hAut(X)$ is 1-connected, but far from an equivalence. If there is torsion in the fundamental group, there may be self-equivalences with non-trivial Whitehead torsion, which thus never lift to homeomorphisms of compact models. I suspect that if you also cross by $R^n$, there are related obstructions for higher $\pi_k$. Maybe $R^\infty$ helps. – Ben Wieland Jan 25 '11 at 4:12