Subnormal not implies conjugate-permutable

Property-theoretic statement
The subgroup property of being a subnormal subgroup does not imply the subgroup property of being conjugate-permutable.

Verbal statement
It is possible to have a subnormal subgroup in a group that is not conjugate-permutable. In fact, it is possible to have a $$k$$-subnormal subgroup that is not conjugate-permutable for any $$k \ge 3$$.

Related facts

 * 2-subnormal implies conjugate-permutable

Example of the dihedral group
Consider a dihedral group of order $$2^n, n \ge 4$$. This is the semidirect product of a cyclic group of order $$2^{n-1}$$ and a cyclic group of order two, where the latter acts on the former by the inverse map. Here, the cyclic subgroup of order two is a $$(n-1)$$-subnormal subgroup. We claim that this subgroup is not conjugate-permutable for $$n \ge 4$$.

To see this, let us make the description of the dihedral group explicit:

$$D_{2^n} := \langle a,x \mid a^{2^{n-1}} = x^2 = e, xax^{-1} = a^{-1} \rangle$$

We claim that the two-element subgroup $$\{ x, e \}$$ is not conjugate-permutable. Indeed, observe that:

$$axa^{-1} = xa^{-2}$$.

So the subgroup $$\{ xa^{-2}, e \}$$ is a conjugate to the original subgroup. On the other hand, for these two subgroups to permute, we require that $$x$$ commutes with $$a^2$$, which is not true since:

$$xa^2x^{-1} = a^{-2} \ne a^2$$

for $$n \ge 4$$.