Derived subgroup not is quasiautomorphism-invariant

Statement
It is possible to have a group $$G$$ such that the commutator subgroup (or derived subgroup) $$[G,G]$$ is not a quasiautomorphism-invariant subgroup, i.e., there exists a fact about::quasiautomorphism $$\sigma$$ of $$G$$ such that $$\sigma(H) \ne H$$.

Related facts

 * Characteristic not implies quasiautomorphism-invariant
 * Quasiautomorphism-invariant not implies 1-automorphism-invariant
 * Center is quasiautomorphism-invariant
 * Center not is 1-automorphism-invariant

Proof
Suppose $$p$$ is a prime number greater than $$3$$. Let $$G$$ be the group isomorphic to the inner automorphism group of the wreath product of two groups of order $$p$$. $$G$$ is a group of order $$p^p$$ with an elementary abelian normal subgroup $$N$$ of order $$p^{p-1}$$, an element of order $$p$$ acting on it from outside, and every non-identity element of $$G$$ has order $$p$$.

Consider the commutator subgroup $$H = [G,G]$$. $$H$$ is a group of order $$p^{p-2}$$, contained inside the elementary abelian normal subgroup.

we can construct a quasiautomorphism $$\sigma$$ of $$G$$ that does not preserve $$H$$ as follows: the restriction of $$\sigma$$ to $$N$$ is an automorphism of $$N$$ that fixes the center $$Z(G)$$ (which is cyclic of order $$p$$) but does not send $$H$$ to itself, and $$\sigma$$ fixes every element of $$G$$ outside $$H$$. Note that we need $$p > 3$$ to ensure that $$H$$ is strictly bigger than $$Z(G)$$, which is necessary to be able to construct a $$\sigma$$ with the desired specifications.