Characteristic not implies injective endomorphism-invariant in finitely generated abelian group

Statement
It is possible to have a finitely generated abelian group $$G$$ and a subgroup $$H$$ of $$G$$ such that $$H$$ is a characteristic subgroup of $$G$$ but is not an injective endomorphism-invariant subgroup of $$G$$, i.e., there exists an injective endomorphism $$\sigma$$ of $$G$$ such that $$\sigma(H)$$ is not contained in $$H$$.

Related facts

 * Characteristic not implies injective endomorphism-invariant
 * Center not is injective endomorphism-invariant
 * Characteristic not implies fully invariant in finite abelian group

Proof
Suppose $$G$$ is the direct product of the group of integers $$\mathbb{Z}$$ and the cyclic group $$\mathbb{Z}/2\mathbb{Z}$$. Suppose $$H$$ is the subgroup generated by the element $$(2,1)$$.


 * $$H$$ is a characteristic subgroup of $$G$$: Let $$A = \{ x \mid \exists y, 2x = 4y \}$$, $$B = \{ x \mid \exists y, x = 2y \}$$,, and $$C = \{ x \mid \exists y, 2x = 8y \}$$. Then, the set $$A \setminus (B \cup C)$$ is the two-element set $$\{ (2,1), (-2,1) \}$$. Both of these are generators of the subgroup $$H$$. Hence, $$H$$ is a characteristic subgroup of $$G$$.
 * $$H$$ is not an injective endomorphism-invariant subgroup of $$G$$: Consider the injective endomorphism $$\sigma$$ defined by $$\sigma((x,y)) = (2x,y)$$. Then, $$\sigma(H)$$ is not contained in $$H$$, because $$\sigma((2,1)) = (4,1)$$, which is not in $$H$$.