Automorph-conjugacy is normalizer-closed

Statement
Suppose $$G$$ is a group and $$H$$ is an automorph-conjugate subgroup of $$G$$, then $$N_G(H)$$ (the normalizer of $$H$$ in $$G$$) is also an automorph-conjugate subgroup of $$G$$.

Generalizations

 * Automorphism-based relation-implication-expressible implies normalizer-closed: This is a general formalism result stating that any subgroup property that can be expressed in a given format is closed under taking normalizers.

Proof idea
The key idea behind the proof is that taking the normalizer commutes with automorphisms. In other words, for any automorphism $$\sigma$$, we have $$\sigma(N_G(H)) = N_G(\sigma(H))$$.

Proof details
Given: A group $$G$$, an automorph-conjugate subgroup $$H$$.

To prove: $$N_G(H)$$ is an automorph-conjugate subgroup: given any automorphism $$\sigma$$ of $$G$$, $$N_G(H)$$ and $$\sigma(N_G(H))$$ are conjugate.

Proof: $$\sigma(N_G(H)) = N_G(\sigma(H))$$. By assumption, $$H$$ is automorph-conjugate, so there exists a $$g \in G$$ such that $$\sigma(H) = gHg^{-1}$$. Thus, $$\sigma(N_G(H)) = N_G(gHg^{-1}) = gN_G(H)g^{-1}$$, and thus, $$N_G(H)$$ and $$\sigma(N_G(H))$$ are conjugate by $$g$$.