Let $G$ be a finitely generated group. Then the automorphism group $\text{Aut}(G)$ of $G$ need not be finitely generated.
However, there are classes of f.g. groups for which the automorphism group will always be f.g., such as polycyclic groups: it is a result by Auslander from 1969 that the automorphism group of a polycyclic group is even finitely presented.
Let $\text{Aut}^0(G) := G$, and for $n \geq 1$ let $\text{Aut}^n(G)$ be defined inductively as $\text{Aut}(\text{Aut}^{n-1}(G))$.
My question is: what are some examples of groups $G$ such that $\text{Aut}^n(G)$ is f.g. for all $n \geq 0$?
Now, while $\text{Aut}(G)$ is f.g. if $G$ is polycyclic by the above, the automorphism group of a polycyclic group need not be polycyclic, as far as I am aware, which would indicate that it is at least conceivable that $\text{Aut}^n(G)$ need not be f.g. for all $n \geq 1$ when $G$ is polycyclic. But I do not know of any counterexamples in this class.
Note that any finite group satisfies the question. Also, if $G$ is taken as $\mathbb{Z}$, then its automorphism group is $C_2$. Are there any infinite examples where every automorphism group is infinite f.g.?
If $G$ is a complete group, that is, with trivial center and no outer automorphisms, then $$G\simeq \text{Aut}(G)$$ as every automorphism is the conjugation by some element, and the map $g\mapsto g\cdot g^{-1}$ has a trivial kernel.
For such a group, the tower is constant, so all you need is to find a finitely generated, infinite, complete group. You will find one in Victor S. Guba's A finitely generated complete group, 1987, DOI:10.1070/im1987v029n02abeh000969.