TITLE: Relation between groups and classifying spaces
QUESTION [9 upvotes]: 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?

REPLY [2 votes]: I thought about the case when $G$ is a free group. I believe that in this case $H_2(BG/G)$ is the second exterior power of $G^{ab}$ (which is also $H_2(BG^{ab})$), and $H_3(BG/G)$ is the kernel of
$$
\oplus (C\otimes C)\to G^{ab}\otimes G^{ab},$$
where $C$ ranges over representatives of conjugacy classes of maximal cyclic subgroups of $G$ (the cokernel of this same map is yet another description of $H_2$), and $H_n(BG/G)$ is trivial for all $n>3$.
EDIT  A useful observation is that if $H$ is a subgroup of $G$ then the fixed point set $(BG)^H$ is $B(Z_GH)$, where $Z_GH$ is the centralizer of $H$ in $G$.