Subgroup of abelian group not implies abelian-extensible automorphism-invariant

Statement
It is possible to have an abelian group $$G$$ and a subgroup $$H$$ of $$G$$ such that $$H$$ is not an abelian-extensible automorphism-invariant subgroup of $$G$$, i.e., there exists an abelian-extensible automorphism $$\sigma$$ of $$G$$ such that $$\sigma(H)$$ is not equal to $$H$$.

Facts used

 * 1) uses::Divisible abelian group implies every automorphism is abelian-extensible

Integers in rationals
Let $$\! G$$ be the additive group of rational numbers, i.e., the abelian group $$(\mathbb{Q},+)$$, $$\! H$$ be the subgroup $$\mathbb{Z}$$ of integers, and $$\! \sigma$$ be the automorphism of $$G$$ given by $$x \mapsto x/2$$.

Then:


 * By fact (1), $$\sigma$$ is abelian-extensible, because $$G$$ is a divisible abelian group.
 * $$\sigma$$ does not send $$H$$ to itself.

Rationals in two copies
Let $$\! G$$ be the direct sum of two copies of the rational numbers, i.e., the abelian group $$\mathbb{Q} \oplus \mathbb{Q}$$, let $$H$$ be the first direct factor, and $$\! \sigma$$ be the automorphism of $$G$$ that interchanges the two coordinates. Then:


 * $$G$$ is a divisible abelian group, so by fact (1), $$\sigma$$ is abelian-extensible.
 * $$\sigma$$ sends $$H$$ to the other direct factor, so $$\sigma$$ does not preserve $$H$$.