Is every connected space equivalent to some B(Aut(X))? Given a connected space $B$, is there always some space $X$ with $B \simeq \mathbf{B}(\mathrm{Aut}(X))$?
Here by space I mean simplicial set, by $\mathrm{Aut}(X)$ I mean the simplicial monoid of auto-equivalences of $X$ (not just strict automorphisms), and by $\mathbf{B}$ I mean the classifying space of this; but I’d also be interested in answers for other reasonable interpretations of these terms.
Edit: A little background — this simply arose out of curiosity, not out of any desired application.  Most of the people I’ve mentioned it to have a strong first impulse that the answer should be “no”, but none of us have been able to substantiate this.
 A: 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, $B \simeq BG \simeq B\text{Aut}_B(E)$.
A: Here is the $1$-type case. I assume all spaces are of the homotopy type of CW. Let me write $haut(X)$ (resp. $haut_*(X)$) for the monoid of self-equivalences (resp. pointed ones) to avoid posible confusion with the group-theoretic notation. These spaces have the correct homotopy type by our assumption.
From All Groups are Outer Automorphism Groups of Simple Groups by Droste,
Giraudet and Göbel, for every discrete group $G$ we can write $G\cong out(S)$ where $S$ is simple. Recall the exact sequence $0\to Z(S)\to S\to aut(S)\to out(S)\to 0$. Since  (thanks to Ricardo Andrade for pointing this) the $S$ appearing in the theorem is in fact non-abelian, $Z(S)=0$.
Looping down the universal fibration with fiber $BS$ we have the homotopy principal fiber sequence
$S\to haut_*(BS)\to haut(BS)$ and thus $haut_*(BS)//S\simeq haut(BS)$. But $haut_*(BS)\simeq aut(S)$ and $S\to aut(S)$ is injective so the homotopy quotient $haut_*(BS)//S$ is the (ordinary) quotient $aut(S)/S\cong out(S)\cong G$.
Hence, $G\simeq haut(BS)$.
