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 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 $Z \simeq Y$.