Let $G$ be a nonabelian group, with classifying space $BG$.  

**Motivation:** We can compute its homology, $H\_\ast(BG)=H\_\ast(G)$. It would be nice to see some equivariant computations, like $H\_\ast^G(BG)$ where $G$ acts by conjugation, but I need a particular model of $BG$ to work with. In any case, if $G$ acted freely it would reduce to computing the homology of $BG/G$. Although this action won't be free, I still wonder what $BG/G$ looks like.


*Preferred Explicit model:* Let $EG$ be the weakly contractible space (constructed simplicially using the elements of $G$), and consider the action of $G\times G$ on $G$ by $(g\_1,g\_2)\cdot g=g\_1gg\_2^{-1}$. Then we recover the classifying space $BG$ as $EG/G$ (with $G=G\times\lbrace 1\rbrace$), and we get the space $BG/G$ as $EG/(G\times G)$.  

[[Edit]: As pointed out, the outcome will depend on the model. If this isn't a "good" model, then I'll settle for a better one! (Although I would want to understand this model).


So I have this space $BG/G$, dividing out the classifying space by the conjugation $G$-action. Here is where I get some big help: By the Kan-Thurston theorem, there exists a group $K$ such that $BG/G$ and $BK$ have the same homologies.


**What can $K$ be?** (Note that if $G$ were abelian then we'd trivially have $K=G$).  

**Is there a deeper connection between these two spaces?**



Relation between groups and classifying spaces
What you should take as a model is the homotopy quotient $EG \times\_G BG$. From the homotopy sequence of the fibration $EG \times\_G BG \to BG$ (projection on first factor), you get that 
$EG \times\_G BG$ is aspherical and a short exact sequence 


$$
1 \to \pi\_1 (BG) \to \pi\_1 (EG \times\_G BG) \to \pi\_1 (BG) \to 1.
$$


Since the action of $G$ on $BG$ has a fixed point, this sequence (which is completely natural in $G$) is split. The induced action of the base $G$ on the fibre $G$ is by conjugation.
So: $EG \times\_G BG \cong BK$; $K = G \ltimes\_{ad} G$, with the conjugation action. 


EDIT: Tom Goodwillie pointed out that $G \ltimes\_{ad} G \cong G \times G$.


