Nilpotent not implies ACIC

Statement
A nilpotent group need not be ACIC: in a nilpotent group, every automorph-conjugate subgroup need not be characteristic.

Example of a non-Abelian group of prime-cubed order
Let $$p$$ be an odd prime. Consider the group $$P$$ of order $$p^3$$ obtained as a semidirect product of a cyclic group of order $$p^2$$ with a cyclic group of order $$p$$. In other words, $$P$$ is given by:

$$\langle a,b \mid a^{p^2} = b^p = e, ab = ba^{p+1} \rangle$$

Then, the cyclic group $$C$$ of order $$p$$ generated by $$b$$ is an automorph-conjugate subgroup that is not characteristic.

$$C$$ is clearly not characteristic because it is not normal, as it is the non-normal part in a semidirect product.

To see that it is automorph-conjugate, observe that any automorphism must send it to another cyclic group of order $$p$$. Now, all cyclic groups of order $$p$$ lie inside the subgroup $$\Omega_1(P)$$, which is elementary Abelian of order $$p^2$$ generated by $$b$$ and $$a^p$$. There are exactly $$p+1$$ subgroups inside this of order $$p$$, and one of them, the center (generated by $$a^p$$) is characteristic. Of the other $$p$$ subgroups, none is normal in $$P$$, and hence, since the size of any conjugacy class of subgroups is a multiple of $$p$$, they are all conjugate. Thus, $$C$$ is an automorph-conjugate subgroup.