Image-closed characteristic not implies fully invariant

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
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.