Suppose that $G$ is a finite group with a normal subgroup $N$. Assume in addition that $\beta$ is an automorphism of group $G/N$. My question is :
Under which conditions we can assure that there is a $\alpha \in Aut(G)$ satisfying two below conditions :
(i) $\alpha (N)=N$
(ii) $\bar \alpha = \beta$, where $\bar \alpha \in Aut(G/N)$ is the automorphism induced by $\alpha$ on $G/N$ with $\bar \alpha(Ng):= N\alpha(g)$
Edition: I know that this might be too general. So for a restriction consider the following question: Apart from family of direct products, which families of groups are known to have a normal subgroup $N$ for which most of $Aut(G/N)$'s members have the above property?
The question is too general to be answered for all quotient groups $G/N$ and $\alpha$. And there is probably no characterization of those $G$ which have this property for all $N$ and $\alpha$.
At least, we easily observe that $\beta$ may be lifted to $G$ when $\beta$ is an inner automorphism of $G/N$. Conversely, if $Q$ is a finite group such that every automorphism $\beta$ of $Q$ may be lifted along any surjective homomorphism $G \to Q$, then $\beta$ is an inner automorphism. The proof is a bit involved, you can find it here.