Automorph-conjugacy is transitive

This article gives the statement, and possibly proof, of a subgroup property (i.e., automorph-conjugate subgroup) satisfying a subgroup metaproperty (i.e., transitive subgroup property)
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Statement

Suppose $H\le K\le G$ are groups such that $H$ is an automorph-conjugate subgroup of $K$, and $K$ is an automorph-conjugate subgroup of $G$. Then, $H$ is an automorph-conjugate subgroup of $G$.

Proof

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Given: Groups $H\le K\le G$ such that $H$ is an automorph-conjugate subgroup of $K$ and $K$ is an automorph-conjugate subgroup of $G$. An automorphism $\sigma$ of $G$.

To prove: There exists $x\in G$ such that $\sigma (H)=xH{x}^{-1}$

Proof:

Step no.	Assertion/construction	Given data used	Previous steps used	Explanation
1	$\sigma (H)\le \sigma (K)$	$H\le K$		given-direct
2	There exists $g\in G$ such that $gK{g}^{-1}=\sigma (K)$	$K$ is an automorph-conjugate subgroup of $G$ $\sigma$ is an automorphism of $G$		given-direct
3	Denote by $c_g$ the map $u\mapsto gu{g}^{-1}$. Then $c_g^{-1}\circ \sigma$ is an automorphism of $G$ that restricts to an automorphism of $K$.		Step (2)	direct from the step
4	There exists $k\in K$ such that $(c_g^{-1}\circ \sigma )(H)=kH{k}^{-1}$.	$H$ is an automorph-conjugate subgroup of $K$	Step (3)	Step-given combination direct
5	Setting $x=gk$, we get that $\sigma (H)=xH{x}^{-1}$		Step (4)	Simple algebraic manipulation gives $\sigma (H)=c_g(kH{k}^{-1})$ $=gkH{k}^{-1}{g}^{-1}=(gk)H(gk{)}^{-1}$