Characteristicity is not finite direct power-closed

Statement with symbols
It is possible to have a group $$G$$ and a characteristic subgroup $$H$$ of $$G$$ such that in the direct square $$G \times G$$, the corresponding direct square $$H \times H$$, viewed as a subgroup, is not a characteristic subgroup.

Related facts

 * Full invariance is finite direct power-closed, hence, fully invariant implies finite direct power-closed characteristic

Notion of finite direct power-closed characteristic
A subgroup $$H$$ of a group $$G$$ is termed a finite direct power-closed characteristic subgroup if for any natural number $$n$$, $$H^n$$ is characteristic in $$G^n$$.

Proof
Suppose $$G = \mathbb{Z}_8 \times \mathbb{Z}_2$$, and $$H$$ is the subgroup of order four given by:

$$H := \{ (2,1), (4,0), (6,1), (0,0) \}$$

Note that this subgroup is characteristic but not fully invariant, and is a standard example of the fact that characteristic not implies fully invariant in finite abelian group.

Then, we claim that $$H \times H$$ is not a characteristic subgroup inside $$G \times G$$.