$\mathrm{Οut}(G) = [BG, BG]$?

Question

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

Answer 1

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 $$[BG, BH] \cong \operatorname{Hom}(G,H)/\operatorname{Inn}(H).$$

This is mentioned in Martino's Classifying spaces and their maps [MR1349123], where this result is attributed to Hurewicz.

Answer 2

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.