# $\mathrm{Οut}(G) = [BG, BG]$?

Let $$G$$ be a finite group and $$BG$$ its classifying space. Let $$[BG, BG]$$ denote the set of self-maps of $$BG$$ up to homotopy equivalence. Automorphisms of $$G$$ give such self-maps, and inner automorphisms are homotopy equivalent to the identity. I am looking for an answer, hopefully with a reference if positive, to the following question:

Is it true that $$\mathrm{Aut}(G)/\mathrm{Inn}(G) \equiv [BG, BG]$$?

Presumably you want all group endomorphisms of $$G$$, instead of just automorphisms. Then it is true that if $$G$$ and $$H$$ are finite groups, then $$[BG, BH] \cong \operatorname{Hom}(G,H)/\operatorname{Inn}(H).$$
The answer's no : take any nontrivial $$G$$ for which $$Out(G) = 1$$, e.g. $$G=\mathbb{Z/2}$$ or $$\mathfrak{S}_n, n \neq 6$$.
Then the isomorphism would imply $$[BG,BG] = 1$$, which would imply that the identity map is nullhomotopic, in other words $$BG$$ would be contractible, which is of course absurd.