Cofactorial automorphism-invariance is not transitive

This article gives the statement, and possibly proof, of a subgroup property (i.e., cofactorial automorphism-invariant subgroup) not satisfying a subgroup metaproperty (i.e., transitive subgroup property).
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Statement

It is possible to have groups $H \leq K \leq G$ such that $H$ is a cofactorial automorphism-invariant subgroup of $K$ and $K$ is a cofactorial automorphism-invariant subgroup of $G$ , but $H$ is not a cofactorial automorphism-invariant subgroup of $G$ .

The key observation here is that the set of prime divisors of the order of $K$ may be a proper subset of the set of prime divisors of the order of $G$ . Shrinkage in the set of prime divisors is necessary to construct a counterexample.

Facts used

p-automorphism-invariant not implies characteristic

Proof

By Fact (1), we can find a prime number $p$ and $p$ -groups $H \leq K$ such that $H$ is $p$ -automorphism-invariant in $K$ but not characteristic in $K$ . In particular, we can find an automorphism of $K$ of order $p^{'}$ relatively prime to $p$ such that $H$ is not invariant under that automorphism. Define $G$ as the semidirect product of $K$ by the cyclic group generated by that automorphism. Then:

$H$ is cofactorial automorphism-invariant in $K$ : $p$ is the only prime dividing the order of $K$ , and $H$ is invariant under $p$ -automorphisms.
$K$ is cofactorial automorphism-invariant in $G$ : In fact, $K$ is a normal Sylow subgroup of $G$ and hence a characteristic subgroup of $G$ . So it is invariant under all automorphisms of $G$ .
$H$ is not cofactorial automorphism-invariant in $G$ : It is not invariant under the conjugation operation by the generator of the cyclic $p^{'}$ -group.

Concretely, using the minimal example provided in the proof of Fact (1), we obtain that for any odd prime, we can construct $H, K, G$ to have orders $p^{3}, p^{6}, 2 p^{6}$ respectively.