Is it true that if $\mathrm{Aut}(G)$ is nilpotent then $G$ is also nilpotent? Is it true that if automorphism group of $G$ is nilpotent then $G$ is also nilpotent?
 A: Yes, if the automorphism group of $G$ is nilpotent, then so is the subgroup of inner automorphisms, which is isomorphic to $G/Z(G)$. But if $G/Z(G)$ is nilpotent, then so is $G$ (standard exercise).
To answer the comment: The converse does not hold. For example, if we take the abelian (and thus nilpotent) group $\mathbb{Z}/2\times \mathbb{Z}/2$ then the automorphism group is isomorphic to $S_3$ which is not nilpotent. If we take some larger still but similar examples, ie $(\mathbb{Z}/p)^n$ for a prime $p$ and any natural number $n$, we get that the automorphism group is isomorphic to $GL_n(\mathbb{F}_p)$ which is never nilpotent (unless $n = 1$) and in fact only solvable for small values of $p$ and $n$.
A: Yep. Also if aut(G) be a nilpotent of dgree n the G be a nilpotent of degree n+1. Since
Let G be a group. Let  Aut(G) be nilpotent then G is nilpotent also. Since
$\frac{G}{Z(G)}‎\cong‎ Inn(G)‎\unlhd‎  Aut(G)$
Since Aut(G) is nilpotent then $\frac{G}{Z(G)}$ is nilpotent. We have
$‎\gamma‎_n(\frac{G}{Z(G)})=Z(G)$
We can write
$‎\gamma‎_n(\frac{G}{Z(G)})=\frac{\gamma‎_n (G)Z(G)}{Z(G)}=Z(G)$
So, $\gamma‎_n (G)Z(G)‎\subseteq‎ Z(G)$ and $\gamma‎_n (G)‎\subseteq Z(G)$. Then G is nilpotent and $\gamma‎_{n+1} (G)‎=e$
