Cocentral not implies amalgam-characteristic

Statement
It is possible to have a group $$G$$ and a cocentral subgroup $$H$$ of $$G$$ (i.e., $$HZ(G) = G$$) such that $$H$$ is not a characteristic subgroup in the amalgamated free product $$L := G *_H G$$.

Similar facts

 * Normal not implies amalgam-characteristic
 * Direct factor not implies amalgam-characteristic
 * Characteristic not implies amalgam-characteristic

Opposite facts

 * Finite normal implies amalgam-characteristic
 * Central implies amalgam-characteristic
 * Normal subgroup contained in hypercenter is amalgam-characteristic

Example of the free group
Let $$F$$ be a free group on two generators and $$\mathbb{Z}$$ be the group of integers. Let $$G = F \times \mathbb{Z}$$ and $$H = F \times \{ 0 \}$$ be the embedded first direct factor. Note that $$HZ(G) = G$$ since the second direct factor is central. So, $$H$$ is a cocentral subgroup. We have:

$$L = (F \times \mathbb{Z}) *_{F \times \{ 0 \}} (F \times \mathbb{Z}) = F \times (\mathbb{Z} * \mathbb{Z}) \cong F \times F$$.

Thus, $$L$$ is a direct product of two copies of the free group on two generators, and moreover, the embedded subgroup $$H$$ in $$L$$ is simply $$F \times \{ e \}$$, the first embedded direct factor. This is not a characteristic subgroup in $$L$$, because there exists an exchange automorphism swapping the two direct factors of $$L$$.