Characteristic not implies isomorph-free in finite group

Statement with symbols
There exists a finite group $$G$$ and a characteristic subgroup $$H$$ of $$G$$ such that $$H$$ is not an isomorph-free subgroup of $$G$$. In other words, there exists another subgroup $$K$$ of $$G$$ that is isomorphic to $$H$$.

Related facts

 * Characteristic not implies isomorph-containing
 * Characteristic not implies sub-isomorph-free in finite
 * Characteristic not implies isomorph-normal in finite
 * Characteristic not implies sub-(isomorph-normal characteristic) in finite
 * Characteristic not implies injective endomorphism-invariant

Example of the dihedral group
Let $$G$$ be the dihedral group of order eight, given as follows, where $$e$$ denotes the identity element of $$G$$:

$$G = \langle a,x \mid a^4 = x^2 = e, xax = a^{-1} \rangle$$.

Let $$H$$ be the center of $$G$$. $$H$$ is a subgroup of order two generated by $$a^2$$.


 * $$H$$ is characteristic.
 * $$H$$ is not isomorph-free: The subgroup $$\langle x \rangle$$ of $$G$$ is isomorphic to $$H$$.