Hereditarily characteristic not implies cyclic in finite

This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties, when the big group is a finite group. That is, it states that in a finite group, every subgroup satisfying the first subgroup property (i.e., hereditarily characteristic subgroup) need not satisfy the second subgroup property (i.e., cyclic characteristic subgroup)
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Statement

It is possible to have a finite group $G$ , and a subgroup $K$ of $H$ such that:

$K$ is a Hereditarily characteristic subgroup (?) of $G$ : For every subgroup $H$ of $K$ , $H$ is a characteristic subgroup of $G$ .
$K$ is not a cyclic group.

Related facts

Finite group implies cyclic iff every subgroup is characteristic: This just states that we cannot get a counterexample when $K=G$ .
Upward-closed characteristic not implies cyclic-quotient in finite

Proof

Further information: nontrivial semidirect product of Z4 and Z4, subgroup structure of nontrivial semidirect product of Z4 and Z4

Consider the semidirect product of cyclic group:Z4 and cyclic group:Z4 where the generator of the acting group acts via the inverse map on the other group. In other words, it has the presentation:

$G:=\langle x,y\mid x^{4}=y^{4}=e,yxy^{-1}=x^{3}\rangle$ .

Let $K$ be the subgroup $\langle x^{2},y^{2}\rangle$ . Then, $K$ is a Klein four-group but every subgroup of $K$ is characteristic in $G$ . We explain here why each of the subgroups of order two is characteristic:

The subgroup $\langle x^{2}\rangle$ is the derived subgroup.
The subgroup $\langle y^{2}\rangle$ is the subgroup generated by the unique element that is a square but not a commutator.
The subgroup $\langle x^{2}y^{2}\rangle$ is the subgroup generated by the unique element that is a product of squares but not a square.

$K$ itself is $\mho ^{1}(G)$ and is characteristic in $G$ , and the trivial subgroup is characteristic in $G$ . Thus, all subgroups of $K$ are characteristic in $G$ .