Characteristic not implies fully invariant in finite abelian group

Statement
In a finite abelian group, a characteristic subgroup need not be a fully invariant subgroup.

Related facts

 * Characteristic equals fully invariant in odd-order abelian group
 * Characteristic not implies fully invariant in finitely generated abelian group
 * Characteristic equals verbal in free abelian group

Proof
Let $$G$$ be the direct sum of the infinite cyclic group and the cyclic group of order two:

$$G := \mathbb{Z}/8\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$$.

Let $$H$$ be the cyclic subgroup of $$G$$ generated by $$(2,1)$$.

The subgroup is characteristic
Set:

$$A := \{g \in G \mid \exists x \in G, 4x = 2g \} = \{ (2r,s) \mid r \in \mathbb{Z}/8\mathbb{Z}, s \in \mathbb{Z}/2\mathbb{Z} \}$$

and

$$B := \{g \in G \mid \exists x \in G, g = 2x \} = \{ (2r,0) \mid r \in \mathbb{Z}/8\mathbb{Z} \}$$.

and:

$$C := \{ g \in G \mid \exists x \in G, 8x = 2g \} = \{ (4r,s) \mid r \in \mathbb{Z}/8\mathbb{Z}, s \in \mathbb{Z}/2\mathbb{Z} \}$$.

Thus, we have:

$$D := A \setminus (B \cup C) = \{ (2r,1) \mid r \in \mathbb{Z}/8\mathbb{Z}, r \notin 2\mathbb{Z}/8\mathbb{Z} \}$$.

Clearly, any automorphism of $$G$$ sends $$A$$ to itself, sends $$B$$ to itself, and sends $$C$$ to itself. Thus, any automorphism of $$G$$ sends $$A \setminus (B \cup C)$$ to itself. Thus, any automorphism of $$G$$ sends $$D$$ to itself. Note that $$D$$ comprises precisely those elements of $$H$$ that have the second coordinate equal to $$1$$: in particular, $$(2,1) \in D \subseteq H$$, so the subgroup generated by $$D$$ equals $$H$$. Thus, any automorphism of $$G$$ preserves $$H$$.

The subgroup is not fully invariant
Consider the map:

$$\pi_1:G \to G, \qquad \pi_1(a,b) = (a,0)$$.

This map is an endomorphism of $$G$$, but the image of $$(2,1)$$ under this map is $$(2,0)$$, which is not an element of $$H$$. Thus, $$H$$ is not fully invariant in $$G$$.