Let $G_0$ be a finite group. Define $G_n = \operatorname{Aut}(G_{n-1})$ for $n \geq 1$.
Does this sequence always eventually end up "constant"? By constant, I mean, for any finite group $G_0$, does there exist an $N \in \mathbb{N}$ such that $G_N \cong G_{N+1} \cong G_{N+2} \cong \dots$?
I'm not sure how to show it generally, nor have I found a counterexample.
I've verified it for groups of order $\leq 15$, as shown below (all such groups appear somewhere in this list).
Write $G_1, \dots, G_n, \to H $ if $\mathrm{Aut}(G_i) \cong H$ for each $i$. If there is no arrow to the right of a group below, it means it is isomorphic to its automorphism group.
$$C_3 \to C_2 \to \{e\}$$
$$C_5 \to C_4 \to C_2 \to \{e\}$$
$$C_7, C_9, C_{14} \to C_6 \to C_2 \to \{e\}$$
$$C_8 \to C_2 \times C_2 \to S_3$$
$$C_{15} \to C_4 \times C_2 \to D_8$$
$$Q_8, A_4 \to S_4$$
$$C_2 \times C_2 \times C_2 \to GL(3,2) \to PGL(2,7)$$
$$C_3 \times C_3 \to GL(2,3) \to C_2 \times S_4$$
$$C_{11} \to C_{10} \to C_4 \to C_2 \to \{e\}$$
$$C_{13} \to C_{12} \to C_2 \times C_2 \to S_3$$
$$\mathrm{Dic}_3, C_2 \times C_6 \to D_{12}$$
$$D_{10} \to GA(1,5)$$
$$D_{14} \to GA(1,7)$$
where $\{e \}$ is the trivial group, $C_n$ is the cyclic group of order $n$, $D_{2n}$ is the dihedral group of order $2n$ $\mathrm{Dic}_n$ is the dicyclic group of order $4n$, $S_n$ (resp. $A_n$) is the symmetric (resp. alternating) group on $n$ letters, $Q_8$ is the quaternion group, $GL(n,q)$, (resp. $GA(n,q)$, $PGL(n,q)$) is the general linear (resp. affine, projective general linear) group of dimension $n$ over $\mathbb{F}_q$.