A finite group, $G$, is nilpotent if its upper central series terminates (at $i \in \mathbb{N})$ with $Z^i(G)=G$, where $Z^i(G)$ is it's $i$-th center which can be described as $\{x \in G\mid \forall y \in G : [x,y]\in Z_{i-1}(G)\}$ where $[x,y]$ is the commutator of $x$ and $y$ in $G$ (in other words, it is the set of elements that commute with one another up to an element of the $(i-1)$-th center).
A finite group, $G$, is non-nilpotent if it is not nilpotent. In this case, we still have, by finiteness of $G$ that the upper central series will terminate after finitely many steps, but now we have an $i \in \mathbb{N}$ at where $Z^i(G)=Z^{i+1}(G)$, and yet $Z^i(G)\neq G$, but rather a proper subgroup of $G$. In this case, since the upper centers are all characteristic and thus normal, we can consider the quotient group, $Q=G/Z^i(G)$. This group must be centerless, else we could take the pullback of $Z(Q)$ under the natural projection to obtain $Z^{i+1}(G)\supset Z^i(G)$ with equality not holding, contradicting our assumption that the upper central series stabilized at $i$. This gives a rather elegant characterization of non-nilpotent finite groups:
A non-nilpotent finite group $G$ can be viewed as an extension of a nilpotent normal subgroup, $U=Z^i(G)$ by a centerless quotient group, $Q$. In the most extreme case, the central series terminates immediately in the trivial group, and you obtain a centerless $G=Q$. This shows that the centerless groups are the primitive elements of the class of non-nilpotent finite groups with the rest being constructed from them by extensions of nilpotent groups by centerless groups.
My question is this: is the following proposition true?: If $G$ is a non-nilpotent finite group, then $\text{Aut}(G)$ is non-nilpotent.
I believe I managed to find reference to the contrapositive of this statement: That if $G$ is a finite group, and $\text{Aut}(G)$ is nilpotent, then so is $G$. But I haven't seen a proof of this, and I would prefer a direct proof of the proposition in question.
If ${\rm Aut}(G)$ is nilpotent then ${\rm Inn}(G) \cong G/Z(G$) is nilpotent and hence $ G$ is nilpotent.