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Let $G$ be a group and $H$ be a subgroup of $G$. When is $\rm{Aut}(H)$ a subgroup of $\rm{Aut}(G)$?

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Generally speaking a question of this type has a shot at a reasonable answer if there is a canonical map $\text{Aut}(H) \to \text{Aut}(G)$ and you want to ask when this map is injective. But in this case there is no such map. Instead there is a canonical map the other way from the subgroup of automorphisms of $G$ preserving $H$ to the group of automorphisms of $H$. So it's not clear to me what you want: do you just want to know when there exists, abstractly, an injection of $\text{Aut}(H)$ into $\text{Aut}(G)$? Do you want this injection to land only in the subgroup of $\text{Aut}(G)$... – Qiaochu Yuan Jan 20 '12 at 6:30
...which preserves $H$ and to respect the action on $H$? (That would require that you could somehow extend automorphisms of $H$ to automorphisms of $G$ in some reasonable way.) There are lots of things you could want and I doubt there is an easy answer to any of the possible forms of this question. – Qiaochu Yuan Jan 20 '12 at 6:31
@Ali Gholamian: Please consider registering in the site; that will make it easier to keep track of your questions. – Arturo Magidin Jan 20 '12 at 16:17
@Ali Gholamian: As above, there is a lot of thing this qestion brings here. You certainly know that $\mathbb{Z}_{6}$ and $\mathbb{Z}_{3}$ know just one group as their Automorphism Group. – Babak S. Jan 21 '12 at 15:14
A partial but also quite trivial answer to this question is: If $\text{Out}(H)=0$, then $\text{Aut}(H)=\text{Inn}(H) \leq \text{Inn}(G) \leq \text{Aut}(G)$. – Nex Oct 17 '15 at 21:47

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