If you are willing to relax your wish from having answer of the form $\text{Aut}(X)$ to the more general group-like topological monoid $\text{Aut}_B(E)$ for a suitable fibration $E \to B$, then the answer is yes. Here $\text{Aut}_B(E)$ is the self-homotopy equivalences of $X$ covering the identity map if $B$. Here's why:

We can assume that $B$ is a CW complex. Let $G \to E \to B$ be a choice of universal principal bundle on $B$, where $G$ is a suitable topological group (here $E$ is contractible, so $B \simeq BG$).

Then $\text{Aut}_B(E) \simeq G$ as topological monoids. Hence, $B \simeq BG \simeq B\text{Aut}_B(E)$.

If you are willing to relax your wish from having answer of the form $\text{Aut}(X)$ to the more general group-like topological monoid $\text{Aut}_B(E)$ for a suitable fibration $E \to B$, then the answer is yes. Here $\text{Aut}_B(E)$ is the self-homotopy equivalences of $E$ covering the identity map if $B$. Here's why:

We can assume that $B$ is a CW complex. Let $G \to E \to B$ be a choice of universal principal bundle on $B$, where $G$ is a suitable topological group (here $E$ is contractible, so $B \simeq BG$).

Then $\text{Aut}_B(E) \simeq G$ as topological monoids. Hence, $B \simeq BG \simeq B\text{Aut}_B(E)$.