Embedding $\operatorname{Aut}(G/Z(G))$ in $\operatorname{Aut}(G)$ The question, which is somewhat open-ended, is this: under which conditions can we guarantee that for a finite group $G$, $\operatorname{Aut}(G/Z(G))$ is isomorphic to a subgroup of $\operatorname{Aut}(G)$?
This is sometimes possible, but not always. It is clearly possible if $G$ has trivial centre or if $G$ is abelian. But it is also possible for $G \cong Q_8$, since $Q_8/Z(Q_8) \cong C_2 \times C_2$ so $\operatorname{Aut}(Q_8/Z(Q_8)) \cong S_3$ whereas $\operatorname{Aut}(Q_8) \cong S_4$. On the other hand, $D_8/Z(D_8) \cong C_2 \times C_2$ again, but $\operatorname{Aut}(D_8) \cong D_8$.
Another case where the embedding I am asking about is possible is when $G/Z(G)$ is a complete group, i.e. it has trivial centre and no outer automorphisms. In that case, $$\operatorname{Aut}(G/Z(G)) \cong G/Z(G) \cong \operatorname{Inn}(G)$$ is a normal subgroup of $\operatorname{Aut}(G)$, but that's not very interesting.
I should, perhaps, clarify that I am mainly interested in what (if anything) we can say from knowledge of $G$ and its subgroup structure alone, without assuming knowledge of $\operatorname{Aut}(G)$. If nothing interesting can be said, however, ignore this restriction.
 A: If $H$ is a perfect group with trivial centre, and $G$ is the Schur covering group of $H$, then $G/Z(G) \cong H$ and ${\rm Aut}(G) \cong {\rm Aut}(H)$.
For example, if $H$ is the simple group ${\rm PSL}(n,q)$ with $n>1$, then with a small number of exceptions (such as ${\rm PSL}(2,9)$ and  ${\rm PSL}(3,4)$), we have $G = {\rm SL}(n,q)$.
To explain the above, let $G$ be any group and write $G = F/R$ with $F$ free. Then an automorphism $\tau$ of $G$ lifts to a homomorphism (not necessarily an automorphism) $\rho:F \to F$ with $\rho(R) \le R$, and so $\rho$ induces a homomorphism $\bar{\rho}:F/[F,R] \to F/[F,R]$.
Since $R/[F,R] \le Z(F/[F,R])$, the restriction, $\sigma$ say, of $\bar{\rho}$ to $[F,F]/[F,R]$ is uniquely determined by $\tau$ - i.e. it is does not depend on the chosen lift to $\rho$.
Also, it is easy to see that applying the same process to $\tau^{-1}$ results in the inverse of $\sigma$ on $[F,F]/[F,R]$, so $\sigma$ is an automorphism. Note that $\sigma$ induces an automorphism of the Schur multiplier $M(G) = ([F,F] \cap R)/[F,R]$ of $G$, and in fact  we get an induced homomorphism ${\rm Aut}(G) \to {\rm Aut}(M(G))$.
The above is true for any group $G$. But if $G$ is perfect, then $[F,F]/([F,F] \cap R) \cong G$, and $[F,F]/[F,R]$ is the unique Schur cover of $G$.               
