Schur-triviality is not characteristic subgroup-closed

This article gives the statement, and possibly proof, of a group property (i.e., Schur-trivial group) not satisfying a group metaproperty (i.e., characteristic subgroup-closed group property).
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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.

Proof

Example of semidihedral group

We take the following:

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