What's the relationship between B Aut(G) and B Aut(BG) for a (discrete) group G? (Some of the notational choices I"m about to make might be iffy; I'm happy to take suggestions for improvements.) 
Let $G$ be a (discrete) group. Think of it as an object in the $2$-category of small categories; then it has an automorphism $2$-group $\text{Aut}(G)$, and this automorphism $2$-group has a classifying space $B \text{Aut}(G)$ given, for example, by taking the geometric realization of the nerve. On the other hand, let $X$ be a space with (weak) homotopy type $BG$. Under mild conditions $\text{Aut}(X)$, the group of homeomorphisms from $X$ to itself, is a topological group, which also has a classifying space $B \text{Aut}(X)$. (Edit: The comments below strongly suggest I should instead be talking about the topological monoid of homotopy equivalences from $X$ to itself, so let's do that instead.) 
Are these classifying spaces (weakly) homotopy equivalent? 
 A: The 2-nerve of the automorphism 2-group of the groupoid $(G\Rightarrow *)$ is indeed homotopy equivalent to $BAut(BG)$.
Indeed, the 
$$
\text{1-nerve of the 2-group of automorphisms of $(G\Rightarrow *)$}
$$
is isomorphic, not just homotopy equivalent, as a group object in simplical sets to the 
$$
\text{simplicial set of automorphisms of $NG$},
$$
where $NG$ denotes the simplicial nerve of the group $G$.
Since $NG$ is a fibrant simplicial set, its simplicial set of self-maps is homotopy equivalent to the topological space of self-maps of $BG:=|NG|$.
That equivalence restricts to an equivalence between the self-maps of $NG$ that are homotopy equivalences and the self-maps of $|NG|$ that are homotopy equivalences. Now, one notes that a self-map of $NG$ is a homotopy equivalence iff it is an isomorphism. Therefore, the
$$
\text{simplicial set of automorphisms of $NG$}
$$
is homotopy equivalent to the topological monoid of
$$
\text{self-homotopy equivalences of $|NG|$}.
$$
Moreover, that equivalence is induced by a map of monoids from the geometric realization of the former to the latter.
To get the equivalence that you asked about, apply $B$ to both sides and use the fact that ``$B$ commutes with geometric realizations'', that is, there is a commutative diagram (up to invertible natural transformation) of functors
$$
\begin{matrix}
\text{group objects in simplcial sets} & \stackrel B \to & \text{simplicial sets}\\
\downarrow |-| & & \downarrow |-| \\
\text{topological groups} & \stackrel B \longrightarrow & \text{topological spaces}
\end{matrix}
$$
