This question is about finite groups.
Let a group $G_0=1$. Suppose $Aut G_1=G_0$. Then $G_1$ is isomorphic to either $1$ or $\mathbb{Z}_2$. To make $|G_1|$ as large as possible, Take $G_1=\mathbb{Z}_2$. Define $G_1=\mathbb{Z}_2$.
Similarly take $AutG_2=G_1$ and let $|G_2|$ take the greatest possible value, so $G_2=\mathbb{Z}_3$.
Take $AutG_{n+1}=G_n$ and let $|G_{n+1}|$ take the greatest possible value.
Questions:
- What are $G_3$ and $G_4$? I really struggle to find them.
- Is the sequence $G_n$ well defined?
- Who can find a necessary and suficient condition on the value of $n$, making $Aut\mathbb{Z}_n=\mathbb{Z}_{n-1}$?
For convenience, I put the answer again here. Thanks Dietrich Burde.
[1]Some Finite Groups Which Are Rarely Automorphism Groups
Theorem: There is no group $G$ such that $Aut(G)$ is abelian of order $p^5$, $p^6$ or $p^7$ for a prime $p$.
Theorem: There is no group $G$ such that $Aut(G)$ has order $p^4$, for odd prime $p$.
More recently, Ban an Yu showed:
Theorem: there is no group $G$ such that $Aut(G)$ is an abelian $p$-group of order $n<p^{12}$, where $p>2$ is a prime, but there is one with $|Aut(G)|=p^{12}$.
For the abelian case: There is no group $G$ such that $Aut(G)\cong C_n$ for $n>1$ odd, see here. A full classification of abelian groups occuring as automorphism groups of finite groups is still not known.