# For a subgroup $H$ of a finite group $G$ , when does $\lvert \operatorname{Aut}(H)\rvert$ divide $\lvert \operatorname{Aut}(G)\rvert$?

Let $$H$$ be a subgroup of a finite group $$G$$. Is it true that $$\lvert \operatorname{Aut}(H)\rvert$$ divides $$\lvert \operatorname{Aut}(G)\rvert$$? What if we also assume $$G$$ is abelian? (I know that $$\lvert \operatorname{Aut}(H)\rvert \space \big| \space \lvert \operatorname{Aut}(G)\rvert$$ if $$G$$ is cyclic).

It's not even true for abelian groups in general. Take $$H=C_2\times C_2$$ as a subgroup of $$G=C_4\times C_2$$. Then $$\lvert \operatorname{Aut}(G)\rvert=8$$, while $$\lvert \operatorname{Aut}(H)\rvert=6$$.