The following might lead to answer to your question (I am posting it as an answer instead of a comment because it takes space). Someone once told me Peter May showed that a fibration is always fiber homotopy equivalent to some fiber bundle (where the fiber can be pretty large).  I don't know how this is proved, and I am not sure if I am even remembering the statement correctly. But here goes...


Assuming it is correct, then to every fibration $E \to B$ with fiber $X$ which is classified by a map $B \to BG(X)$, where $G(X)$ is the monoid of homotopy automorphisms, there is supposed to be a homotopy equivalence $Y \to X$ and a factorization up to homotopy of the classifying map:
$$
B \to B\text{aut}(Y)  \to BG(Y) \simeq BG(X) .
$$
So if we consider the universal (quasi-)fibration 
$$
F \to EG(X) \times_{G(X)} F \to BG(X)
$$
with classifying map the identity $BG(X) \to BG(X)$, it will factorize through $B\text{aut}(Y)$ for some choice of $Y$.  This would show at least that $BG(X)$ is a retract of $B\text{aut}(Y)$.


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*Note added:* the result of Peter May I quoted is more-or-less contained in a paper by Casson and Gottlieb. The idea is this: any finite CW $X$ is homotopy equivalent to a codimension zero
open submanifold $M \subset \Bbb R^n$ for $n$ large (this is given by thickening). 

They show if $n$ is large that 
$$
[B,B\text{diff}(M)] \to [B,BG(M)] = [B,BG(X)]
$$
is surjective, where $B$ is finite dimensional CW and $n$ depends on $\dim X$ and $\dim B$.
This shows that any finite skeleton of $BG(X)$ is a retract of 
$B\text{diff}(M \times \Bbb R^j)$ for suitable choice of $j$. 

Since $B\text{diff}(M) \to BG(M)$ factors through $B\text{homeo}(M)$, we  get the same statement for homeomorphisms. Finally, we can take $j \to \infty$. This will show that
$BG(X)$ is a retract of $B\text{homeo}^{\text{st}}(M)$, where 
$$
\text{homeo}^{\text{st}}(M) = \lim_j \quad \text{homeo}(M\times \Bbb R^j) .
$$
By the way, this last group maps to $\text{homeo}(M \times Q)$ where $M$ is the Hilbert cube. So we get that $BG(X)$ is a retract of $B\text{homeo}(M \times Q)$.