Characteristic not implies amalgam-characteristic

Statement
A characteristic subgroup of a group need not be an amalgam-characteristic subgroup.

Related facts

 * Normal not implies amalgam-characteristic
 * Finite normal implies amalgam-characteristic
 * Central implies amalgam-characteristic
 * Normal subgroup in upper central series member is amalgam-characteristic

Example of the infinite dihedral group
Let $$G = D_\infty$$ be the infinite dihedral group given by:

$$G = \mathbb{Z} \rtimes \mathbb{Z}/2\mathbb{Z}$$.

Let $$H$$ be the subgroup of $$G$$ given as the normal subgroup $$\mathbb{Z}$$. Note that $$H$$ is characteristic in $$G$$, since it is the centralizer of derived subgroup. We have:

$$L = G *_H G = \mathbb{Z} \rtimes (\mathbb{Z}/2\mathbb{Z} * \mathbb{Z}/2\mathbb{Z}) = \mathbb{Z} \rtimes D_\infty = (\mathbb{Z} \times \mathbb{Z}) \rtimes \mathbb{Z}/2\mathbb{Z}$$

Note here that we're using the fact that the infinite dihedral group can be identified with the free product of two copies of the cyclic group of order two, whence either of these takes the role of the acting $$\mathbb{Z}/2\mathbb{Z}$$. The upshot is that we have:

$$L = (\mathbb{Z} \times \mathbb{Z}) \rtimes \mathbb{Z}/2\mathbb{Z}$$

where the action is by the inverse map. Further, $$H$$ is the first embedded direct factor $$\mathbb{Z} \times \{ 0 \}$$ in $$\mathbb{Z} \times \mathbb{Z}$$.

Finally, observe that the coordinate exchange automorphism of $$\mathbb{Z} \times \mathbb{Z}$$ extends to an automorphism of $$L$$, since it commutes with the inverse map, and can therefore be taken to fix the complementary $$\mathbb{Z}/2\mathbb{Z}$$.

Thus, $$H$$ is not a characteristic subgroup of $$L$$.