$\DeclareMathOperator{\Char}{char}$$\DeclareMathOperator{\Aut}{Aut}$ Let $G$ be a finite group, $H$ a subgroup, such that $H \Char G$. If $\phi \in \Aut(H)$ is there an automorphism $\widehat{\phi} \in \Aut(G)$ such that $\widehat{\phi}|_H$ = $\phi$?
In words, if $\phi$ is an automorphism of $H$, is there necessarily an automorphism $\widehat{\phi}$ of $G$ where $\widehat{\phi}$ maps the elements in $H$ to $H$ in exactly the same way?
I am thinking this is false (consider the non-finite example of $\bar{Q}$ and $\mathbb{C}$), but would like a counter example (for the finite case).
As you said, the answer is not always true, not always false.
Let $G$ be the non-abelian $p$-group ($p>2$), given by $$G=C_{p^2}\rtimes C_p=\langle x,y \colon x^{p^2}, y^p, yxy^{-1}=x^{1+p}\rangle .$$ In this group, $\langle x^p,y\rangle\cong C_p\times C_p$ is characteristic subgroup, since it is the only maximal subgroup which has exponent $p$. This characteristic subgroup has lot of automorphisms, take a simple one $$x^p\mapsto y, \mbox{ and } y\mapsto x^p.$$ This automorphism of $H$ can not be extended to an automorphism of $G$, because, $x^p$ is in center of $G$ but $y$ is not.
The example suggests that while looking extension of automorphisms from subgroup to group, we should also care about characteristic subgroups of $G$ which are inside $H$; such subgroups may not be characteristic in $H$, but in bigger group, they may become characteristic.