Image-closed characteristic not implies fully invariant

This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., image-closed characteristic subgroup) need not satisfy the second subgroup property (i.e., fully invariant subgroup)
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Statement

It is possible to have a group $G$ and an image-closed characteristic subgroup $H$ of $G$ such that $H$ is a fully invariant subgroup of $G$.

Proof

Further information: direct product of S3 and Z3

Suppose $G$ is the direct product of the symmetric group of degree three and the cyclic group of order three. Suppose $H$ is the second direct factor of $G$, so $H$ is the cyclic group of order three. We have:

• For any surjective homomorphism $\rho :G\to K$, $\rho (H)$ is a characteristic subgroup of $K$: Note first that $H$ itself equals the center of $G$, hence is characteristic in $G$. For any surjective homomorphism, the kernel is either $H$ or the cyclic subgroup of order three inside the first direct factor. In either case, $\rho (H)$ is characteristic in $K$.
• $H$ is not a fully invariant subgroup of $G$: $H$ is not invariant under the endomorphism of $G$ that has kernel the first direct factor and sends $H$ to the cyclic subgroup of order three inside the first direct factor.