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.



Is every connected space equivalent to some B(Aut(X))?
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*](http://dx.doi.org/10.1112/S0024610701002484) 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, $G\simeq haut(BS)$.


