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 $B\text{aut}(Y)$.