Characteristic not implies injective endomorphism-invariant

Statement with symbols
It is possible to have a group $$G$$ with a characteristic subgroup $$H$$ that is not injective endomorphism-invariant: in other words, every automorphism of $$G$$ sends $$H$$ to itself, but every injective endomorphism of $$G$$ does not send $$H$$ to itself.

Facts used

 * 1) uses::Finitary symmetric group is characteristic in symmetric group
 * 2) uses::Finitary symmetric group is not injective endomorphism-invariant in symmetric group
 * 3) uses::Center is characteristic
 * 4) uses::Center not is injective endomorphism-invariant

Example of the finitary symmetric group
Let $$S$$ be an infinite set. Let $$G = \operatorname{Sym}(S)$$ be the symmetric group on $$S$$, and let $$H = \operatorname{FSym}(S)$$ be the subgroup comprising finitary permutations. Then, $$H$$ is characteristic in $$G$$ (fact (1)) but is not I-characteristic in $$G$$ (fact (2)).

Example of the center
Facts (3) and (4) give another kind of example: the center of a group is always characteristic, but need not be injective endomorphism-invariant.