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 $X$?



Extreme rigidification of homotopy self-equivalences
This is a substantial revision of my original post. It shows that if we replace the "equivalence" Tyler is asking for by a "retract" then the answer is yes.


1. Given a CW space $Y$, we can take $G(Y) =$ the topological monoid of homotopy automorphisms of $Y$. The Borel construction
$$
EG(Y) \times\_{G(Y)} Y \to BG(Y)
$$
is then a quasifibration. Let $U \to BG(Y)$ be the effect of converting it into a fibration.
2. Let $G$ be a topological group with a chosen homotopy equivalence $$BG\simeq BG(Y). $$ For example, we can do what
Tyler does, or we can simply take $\Omega BG(Y)$, where this means the realized Kan loop of the total singular complex of $BG(Y)$.
3. Let $EG \to BG$ be a universal $G$-principal bundle, and set 
$$
Z \quad := \quad \text{pullback}(EG \to BG \simeq BG(Y) \leftarrow U)
$$
Then $Z \subset EG \times U$ inherits a $G$-action and its underlying homotopy type is that of $Y$. Then the Borel construction
$$
EG\times\_G Z \to BG
$$
is a fiber bundle which is weak fiber homotopy equivalent to $U \to BG(Y)$.
4. Step 3 implies that $BG(Y)$ is a retract up to homotopy of $B\text{homeo}(Z)$.
This will imply that $G(Y)$ is a homotopy retract of $\text{homeo}(Z)$ in the $A\_\infty$ sense, with $Z \simeq Y$.


