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.
A (finite) group, $G$, is complete if it is centerless and isomorphic to its automorphism group $G\simeq\text{Aut}(G)$.
Conjecture: Suppose we have a non-nilpotent finite group $G$, whose upper central series terminates at $i>0$, meaning that $G$ is not centerless, and $Z^i(G)=Z^{i+1}(G)$. Let $U=Z^i(G)$, and suppose further that the centerless quotient $Q=G/U$ is complete, meaning $Q\simeq\text{Aut}(Q)$. Lastly, suppose that the upper central series for $\text{Aut}(G)$ terminates at some $i_1\geq i$, that is, the upper central series for $\text{Aut}(G)$ is at least as long as the upper central series for $G$. Note in particularly that $\text{Aut}(G)$ is non-nilpotent, and so has a centerless quotient $Q_1$. Then the claim is that the centerless quotients are isomorphic, $Q_1\simeq Q$.
I came up with the conjecture based on computer searches in GAP looking at the automorphism series for various small finite groups as far as I could construct them in GAP (many eventually reach very large groups and run into issues with memory or the size of the computation). Looking at the upper central series for each term of the automorphism series and comparing how they changed as you step through the automorphism series lead me to propose that the above conjecture holds in general for finite groups. Naturally a counterexample will be enough to disprove the claim, but I didn't find any looking at over 50 different automorphism series, so it may not be easy to find if it exists. I'm not really sure how to approach trying to prove this either. I don't really know enough about the relationship between a group and its automorphism group to leverage the givens in this conjecture productively.
Edit 1: This is a response to David A. Craven's answer:
I took the liberty of constructing the proposed group in GAP for $n=7$. To do this, I needed to put all three of the $G_i$'s into permutation groups so that the direct product operation in GAP would also return a permutation group, so that fast algorithms would be available for subsequent calculations. Specifically I used:
G1:=SymmetricGroup(7);
G2:=Group((1,2,3,4),(1,2)(3,4)); #This is D8
G3:=Group((1,2,3),(4,5,6),(7,8,9)); #This (Z_3)^3
#That G2 and G3 are as claimed can be checked by IdSmallGroup, is should return (8,3) for G2 and (27,5) for G3.
#Now we can call DirectProduct and the result will be a permutation group
G:=DirectProduct(G1,G2,G3);
#Because this is a permutation group, computation of the upper central series is 'easy' for GAP, trying this without doing the above preparations lead to a computation that was taking minutes and still hadn't finished
U:=UpperCentralSeriesOfGroup(G);
From the above I obtained that $G$ has upper central series of length $2$. With $Z(G)\simeq (54,15)$, and $Z^2(G)\simeq (216,151)$ (Where (x,y) is the GAP library id for order x, id y.). GAP StructureDescription for U[1] confirms that $Z^2(G)\simeq\mathbb{Z}_3^3\times D_8$. Next I constructed the quotient and confirmed it was the symmetric group (necessarily $S_7$ by an easy order argument). After a computation that took a few seconds I got the automorphism group, $A1$, but it was not in the form of a permutation group, so I used NiceMonomorphism and then NiceObject to obtain a permutation group isomorphic to the automorphism group. This allows for fast(er) algorithms to used for computing its upper central series, which I found to be of length $2$ with $Z(A1)\simeq (8,5)$ and $Z^2(A1)\simeq (32,46)$ (again using GAP ids for the groups). The size of $A1$ is 905748480 which is twice that predicted in the answer (Which would be 452874240 if it had the structure given in the answer). I was able to get GAP to give me a structure description for $A1$, which was $\mathbb{Z}_2^2\times D_8\times S_7\times \text{PSL}(3,3)$. I suspect the comment that the proposed automorphism group is, in fact, a subgroup of index 2 in the full automorphism group is correct. The quotient $Q1$ is of order 28304640 (and so obviously not isomorphic to $S_7$), GAP gives a structure description of $S_7\times \text{PSL}(3,3)$ for this quotient. (So it is still a counter example for $i_1=i$ case).
Best Answer
I ran a brute-force search on groups of small order using the following code:
Letting it run over the groups of order $\leq 72$ produced three counterexamples:
EDIT: There also exist counterexamples with $i_1 > i$, e.g.