Normalizer of isomorph-conjugate implies isomorph-dominating

Statement
Suppose $$G$$ is a group and $$H$$ is an fact about::isomorph-conjugate subgroup of $$G$$: any subgroup of $$G$$ isomorphic to $$H$$ is also conjugate to $$H$$ within $$G$$. Then, the fact about::normalizer $$N_G(H)$$ of $$H$$ in $$G$$ is isomorph-dominating in $$G$$: if $$K \le G$$ is isomorphic to $$N_G(H)$$, there exists $$g \in G$$ such that $$K \le gN_G(H)g^{-1}$$.

Corollaries

 * Isomorph-conjugacy is normalizer-closed in finite: In a finite group, any isomorph-dominating subgroup is also isomorph-conjugate, so the normalizer of any isomorph-conjugate subgroup is isomorph-conjugate.

Note that the normalizer of an isomorph-conjugate subgroup in an infinite group is not necessarily isomorph-conjugate. For instance, consider the group of integers. The normalizer of the trivial subgroup is the whole group, which is not isomorph-conjugate: there are proper subgroups of the group of integers isomorphic to it.

Proof
(This proof uses the left-action convention).

Given: A group $$G$$, an isomorph-conjugate subgroup $$H$$ of $$G$$.

To prove: If $$K \le G$$ is a subgroup such that $$K \le N_G(H)$$, then there exists $$g \in G$$ such that $$K \le gN_G(H)g^{-1}$$.

Proof: Let $$\sigma:N_G(H) \to K$$ be an isomorphism. Then, $$H$$ and $$\sigma(H)$$ are isomorphic groups. Thus, there exists $$g \in G$$ such that $$gHg^{-1} = \sigma(H)$$.

Since taking normalizer commutes with automorphisms, we get $$gN_G(H)g^{-1} = N_G(\sigma(H))$$. Now, since $$H$$ is normal in $$N_G(H)$$, $$\sigma(H)$$ is normal in $$\sigma(N_G(H)) = K$$. In particular, $$K$$ is contained in the normalizer of $$\sigma(H)$$, so $$K \le N_G(\sigma(H)) = gN_G(H)g^{-1}$$.