Restriction of automorphism to subgroup invariant under it and its inverse is automorphism

Statement
Suppose $$G$$ is a group, $$H$$ is a subgroup, and $$\sigma$$ is an fact about::automorphism of $$G$$ such that both $$\sigma$$ and $$\sigma^{-1}$$ leave $$H$$ invariant. Then, $$\sigma$$ restricts to an automorphism of $$H$$.

Related facts

 * Restriction of automorphism to subgroup not implies automorphism: If $$\sigma$$ is an automorphism of a group $$G$$ and $$H$$ is a subgroup of $$G$$ such that $$\sigma(H) \subseteq H$$. The restriction of $$\sigma$$ to $$H$$ need not be an automorphism of $$H$$.
 * For any group-closed automorphism property (i.e., any property of automorphisms such that for any given group the automorphisms satisfying the property form a group), any subgroup invariant under all automorphisms with the property satisfies the additional condition that the restriction of each such automorphism to the subgroup is an automorphism of the subgroup. A subgroup property that can be expressed this way is termed an auto-invariance property. Examples of this are:
 * The property of being a characteristic subgroup is the invariance property with respect to all automorphisms. The restriction of any automorphism of the whole group to a characteristic subgroup is an automorphism of the subgroup.
 * The property of being a normal subgroup is the invariance property with respect to all inner automorphisms. The restriction of any inner automorphism of the whole group to a normal subgroup is an automorphism of the subgroup.

Proof
Given: A group $$G$$, a subgroup $$H$$, an automorphism $$\sigma$$ of $$G$$ such that $$\sigma(H) \subseteq H$$ and $$\sigma^{-1}(H) \subseteq H$$.

To prove: $$\sigma(H) = H$$ and the restriction of $$\sigma$$ to $$H$$ is an automorphism of $$H$$.

Proof:


 * 1) Since $$\sigma$$ is an automorphism of $$G$$, so is $$\sigma^{-1}$$, and their composite (both ways) is the identity map on $$G$$. In other words, $$\sigma(\sigma^{-1}(g)) = g$$ and $$\sigma^{-1}(\sigma(g)) = g$$ for all $$g \in G$$.
 * 2) By our assumption, the restrictions $$\sigma|_H$$ and $$\sigma^{-1}|_H$$ are both functions from $$H$$ to itself. Further, we have that $$\sigma(\sigma^{-1}(h)) = h$$ and $$\sigma^{-1}(\sigma(h)) = h$$ for all $$h \in H$$. Thus, $$\sigma|_H$$ and $$\sigma^{-1}|_H$$ are two-sided inverses of each other, and are thus both bijections. In particular, $$\sigma(H) = H$$.
 * 3) Finally, since $$\sigma$$ is a homomorphism, so is $$\sigma|_H$$. Thus, $$\sigma|_H$$  is a bijective homomorphism from $$H$$ to itself, and is hence an automorphism of $$H$$.