Are there automorphisms of $H$ which are not restriction of an automorphism of $G$? Let $G$ be a group and $H$ a characteristic subgroup of $G$ (that is, invariant under all automorphisms of $G$).
Let $\phi \in \mathcal{Aut}(H)$, we'll call $\widetilde{\phi}$ an automorphism of $G$ such that

$$\widetilde{\phi}(h) = \phi(h) \quad \forall h \in H$$

Or in other words, such that $\widetilde{\phi}_H = \phi$, where $\widetilde{\phi}_H$ denotes its restriction to $H$.

Can we construct a $\widetilde{\phi}$ for every $\phi \in \mathcal{Aut}(H)$?

Or in other words, are there automorphisms of $H$ which are not restrictions to $H$ of an automorphism of $G$?

It's easy to show that if the answer is no, then there is a bijection $\mathfrak{A}: \mathcal{Aut}(H) \xrightarrow{\sim} \mathcal{Aut}(G)$ such that $\mathfrak{A}(\phi) = \widetilde{\phi}$. Could this fact be used to disprove the claim above by contradiction?
The answer is no. One counterexample is the group $G = {\rm AGL}(1,8)$, which has order $56$. It has a normal (and therefor characteristic) Sylow $2$-subgroup $P$, which is elementary abelian of order $8$.

Now ${\rm Aut}(G)$ is the group ${\rm A \Gamma L}(1,8)$, which is a solvable group of order $168$ and has $G$ as a normal subgroup of index $3$. However, ${\rm Aut}(P)$ is the group ${\rm PSL}(3,2)$, which coincidentally also has order $168$, but it is a nonabelian simple group. In fact only $21$ of the $168$ automorphisms of $P$ extend to automorphisms of $G$.
If $H$ is characteristic in $G$ then every automorphism of $G$ is an automorphism of $H$. Thus, the question reduces to answering whether $A(H) \subset A(G)$ is true. Consider $Gh_1$ and $Gh_2$ where $h_1$ and $h_2$ are in $H$. Then $g_1h_1g_2h_2=g_1g_2'h_1h_2$. This shows that the right-regular representation of $H$ is a subgroup of the right-regular representation of $G$. Thus the holomorph of $H$ is a subgroup of the holomorph of $G$. Since $A(H)$ and $A(G)$ fix the identity in the respective holomorphs, $A(H) \subset A(G)$.

EDIT: The claim that the holomorph of $H$ is a subgroup of the holomorph of $G$ is indeed incorrect, for the normalizer of the right regular representation of $H$ in $S_G$(The symmetric group on elements of $G$) need not necessarily be contained in the normalizer of the right regular representation of $G$ in $S_G$.