Let $C_n$ be the cyclic group of order $n$. Its automorphism group $Aut(C_n)$ is a group of order $\varphi(n)$ isomorphic to $(\mathbb{Z}/n\mathbb{Z})^{\times}$ the multiplicative group of integer modulo $n$. This last group is abelian but not always cyclic, the first non-cyclic being $Aut(C_8) \simeq C_2 \times C_2$. Moreover the iterated automorphism groups $Aut^m(C_n)$ are not always abelian, as $Aut^2(C_8) \simeq S_3$.

By watching the table here for the group structure of $(\mathbb{Z}/n\mathbb{Z})^{\times}$, we cannot expect an easy classification for the set of groups $Aut(C_n)$, but (**Q1**) what about the following set? $$\{Aut^m(C_n) \ | \ n \ge 1, m \ge 0 \}$$

If it is also non-easy, (**Q2**) what about the set of groups $Aut^m(C_n)$ which are isomorphic to $Aut^{m+1}(C_n)$? For $n \le 15$, it is exactly $\{C_1,S_3,D_8 \}$ as shown by the following table:

$$\scriptsize{ \begin{array}{c|c} G &C_1&C_2&C_3&C_4&C_5&C_6&C_7&C_8&C_9&C_{10}&C_{11}&C_{12}&C_{13}&C_{14}&C_{15} \newline \hline Aut(G) &-&C_1&C_2&C_2&C_4&C_2&C_6&C_2^2&C_6&C_4&C_{10}&C_2^2&C_{12}&C_6&C_2 \times C_4 \newline \hline Aut^2(G) &-&-&C_1&C_1&C_2&C_1&C_2&S_3&C_2&C_2&C_4&S_3&C_2^2&C_2&D_8 \newline \hline Aut^3(G) &-&-&-&-&C_1&-&C_1&-&C_1&C_1&C_2&-&S_3&C_1&- \newline \hline Aut^4(G) &- &- &-&-&-&-&-&-&-&-&C_1&-&-&-&- \end{array} }$$

Let $\alpha(n)$ be the smallest $m \ge 0$ such that $Aut^m(C_n) \simeq Aut^{m+1}(C_n)$. Then:

$$\scriptsize{ \begin{array}{c|c} n &1&2&3&4&5&6&7&8&9&10&11&12&13&14&15 \newline \hline \alpha(n) &0&1&2&2&3&2&3&2&3&3&4&2&2&3&2 \end{array} }$$

Surprisingly, for $n<15$ (but not for $n=15$), $\alpha(n)$ is exactly the number of iterations of the Carmichael lambda function $\lambda$ needed to reach $1$ from $n$ (OEIS A185816). (**Q3**) Why?

A specific OEIS sequence has just been created (A331921) providing the value of $\alpha(n)$ for $n<32$; in this case, $\alpha(n) \le 5$ and $\{\mathrm{Aut}^5(C_n) \ | \ n < 32 \} = \{C_1,S_3,D_8,D_{12},\mathrm{PGL}(2,7) \}$. A usual laptop cannot compute $\alpha(n)$ for $n \ge 32$ (you are welcome to contribute to this sequence), we just know that $\alpha(32) \ge 6$. Here is the sequence $|Aut^n(C_{32})|$ for $n \le 6$:

$$\scriptsize{ \begin{array}{c|c}
n &0&1&2&3&4&5&6&7&8 \newline
\hline
|Aut^n(C_{32})| & 2^5&2^4&2^4&2^6&2^{7}3&2^{9}3&2^{11}3&?&? \newline
\hline
\text{IdGroup}(Aut^n(C_{32})) & [32,1]&[16,5]&[16,11]&[64,138]&[384,17948]&[1536,?]&[6144,?]&[?,?]&[?,?]
\end{array} }$$

We can also consider the sequence of $n$ such that $\exists m \ge 0$ with $Aut^m(C_n) \simeq C_1$:

$1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 14, 18, 19, 22, 23, 27, 38, 46, 47, 54, 81, \dots$ (A117729).
It seems to be a (strict) subsequence of A179401, so (**Q4**) why $Aut^m(C_n) \simeq C_1$ $\Rightarrow$ $\varphi^2(n) = \lambda^2(n)$?

It is an open problem whether the sequence $Aut^n(G)$ stabilizes (see here). For a given finite group $G$ there are three possibilities:

- (1): $\exists m \ge 0 $ such that $Aut^{m+1}(G) \simeq Aut^{m}(G)$,
- (2): not (1) and $\exists m \ge 0$ such that $\exists r>0$ with $Aut^{m+r}(G) \simeq Aut^{m}(G)$,
- (3): not (1) and not (2), i.e., $\forall m \ge 0$ and $\forall r>0$ then $Aut^{m+r}(G) \not \simeq Aut^{m}(G)$.

The case (1) means that the sequence $(Aut^m(G))_m$ is constant for $m$ large enough, (2) means that it is periodic non-constant for $m$ large enough, and (3) means never periodic. The existence of finite groups in cases (2) or (3) is open. (**Q5**) Can we rule out the cases (2) and (3) for the cyclic groups?

If not, let redefine $\alpha(n)$ as the smallest $m \ge 0$ such that $\exists r>0$ with $Aut^{m}(C_n) \simeq Aut^{m+r}(C_n)$ if $C_n$ is in cases (1) or (2), and $\infty$ in the case (3).