Schur-triviality is not characteristic subgroup-closed

Statement
It is possible to have a Schur-trivial group $$G$$ and a characteristic subgroup $$H$$ of $$G$$ such that $$H$$ is not a Schur-trivial group.

Example of semidihedral group
We take the following:


 * $$G$$ is particular example::semidihedral group:SD16, which is a Schur-trivial group.
 * $$H$$ is the subgroup particular example::D8 in SD16 inside $$G$$. This is an isomorph-free subgroup, hence is characteristic.
 * $$H$$ is abstractly isomorphic to particular example::dihedral group:D8, which is not Schur-trivial.