Is the statement that $ \operatorname{Aut}( \operatorname{Hol}(Z_n)) \cong \operatorname{Hol}(Z_n)$ true for every odd $n$? Is the statement that $ \operatorname{Aut}(\operatorname{Hol}(Z_n)) \cong  \operatorname{Hol}(Z_n)$ true for every odd $n$? $Hol$ stands here for group holomorph.
This problem appeared, when I stumbled upon the following MO question: https://mathoverflow.net/questions/258886/conditions-for-a-finite-group-to-be-isomorphic-to-its-automorphism-group
OP of that question provided us with the complete list of groups $G$, such that $|G| \leq 506$ and $ \operatorname{Aut}(G) \cong G$. Among those groups there are some looking like holomorphs of all cyclic groups of odd orders up to $23$. Does anybody know, if that pattern continues or is it just a coincidence?
 A: Here's a proof, which is really an expansion of my comment above.
Let $A=\langle a\rangle$ be a cyclic group of order $n$, with $n$ odd.  Let $K=\operatorname{Aut}(A)$, and consider the holomorph $G=A\rtimes K$. Because $n$ is odd, $Z(G)$ is trivial: because the inversion map is in $K$, we see that $C_G(a)=A$, and no element of $A$ is central. Also, since $K$ is abelian, we see $A=[G,G]$.
Now let $\phi$ be an automorphism of $G$. Then $\phi(A)=A$, and so there exists $k\in K$ such that $\phi(a)=a^k$. Now $H=\phi(K)$ is a self-centralizing subgroup of $G$ that is a complement to $A$. Because it's a complement, for every $g\in K$, there's a unique element of the form $a^?g\in H$. In particular, consider $\iota\in K$, the inversion map. If $a^r\iota\in H$, then setting $m=-r(n+1)/2$, it is easy to check that $\iota\in a^mHa^{-m}$. By looking at $[\iota,ga^s]$, we see that $C_G(\iota)=K$, and so $a^mHa^{-m}=K$. Thus $\phi$ acts on $G$ exactly like conjugation by $ka^m$, so $\phi$ is inner. Combined with the triviality of $Z(G)$, we see $G$ is complete.
Edit: I might have glossed over one too many details in the end above. Let $\psi\in\operatorname{Aut}(G)$ be conjugation by $ka^m$, and let $\alpha=\phi\psi^{-1}$. Then we've shown $\alpha$ fixes $A$ pointwise, and $K$ setwise. But then for any $g\in K$, we have
\begin{align}
a^g &= \alpha(a^g)\\
&= a^{\alpha(g)}
\end{align}
and thus $g$ and $\alpha(g)$ are two automorphisms of $A$ with the same action, meaning $\alpha(g)=g$. Thus $\alpha$ fixes $K$ pointwise, and since $G=AK$, $\alpha$ is the identity map.
