Question:
Let $G$ be a finite group. Let $Aut(G)$ be the group of automorphisims of $G$. Consider the group action $\phi:Aut(G)\times G\to G$ where $\phi(\sigma,g)=\sigma(g)$. Assume $G$ has exactly two orbits under the action of $Aut(G)$.
(a) Determine all such $G$, up to isomorphism.
(b) List all cases in which $Aut(G)$ is a solvable group.
Answer:
In the part (a), I showed that these two orbits must be $\{1\}$ and $G\setminus \{1\}$. Thus, if we fix $g\in G\setminus \{1\}$, then for any $g'\in G\setminus \{1\}$, there exists $\sigma\in Aut(G)$ so that $\sigma(g)=g'$. Since automorphisms preserve order, this yields that except $1$, every element has the same order in $G$. Then by using Cauchy's theorem, I proved that $|G|=p^n$ and since every element has the same order, $G$ must be an elementary abelian group, i.e. $G=C_p^n$.
Now in the part (b), I said that as $G=(\mathbb{Z}/p\mathbb{Z})^n$, we can regard $G$ as an $n$-dimensional vector space over $\mathbb{Z}/p\mathbb{Z}$. Then I guess that $Aut(G)=GL_n(\mathbb{Z}/p\mathbb{Z})$ (I am not sure if this part is true). Even if this last part is true, I do not know how to find the cases where $GL_n(\mathbb{Z}/p\mathbb{Z})$ is a solvable group.
Best Answer
(posting an answer since my comment was too long)
Part a)
I'm afraid you're going a bit too quick between the proof that $|G|=p ^n$ and the element all have same order and then the conclusion that $G$ must be an elementary abelian group.
For instance, the Heisenberg group (upper triangular 3x3 matrices with ones on the diagonal) over $\mathbb{F}_p$ has order $p^3$ and all elements have order $p$ for $p \geq 3$.
However, you can still use the hypotheses to reach this conclusion. Indeed, you have proved that $G$ is a $p$-group. It is well known that $p$-group have non trivial center. So there is a central non trivial element and all non trivial elements are conjugate to it. All conjugates of a central element are central and identity is central also. The group is thus equal to its center.
You can then conclude that $G$ is abelian and then elementary because all non-trivial elements have order $p$.
Part b)
An elementary group of order $p^n$ is a vector space over the finite field $\mathbb{F}_p$. If you consider a automorphism, it translates into a vector space homomorphism, which must be injective. This confirms that $Aut(G)=GL_n(\mathbb{F}_p)$.
Determining when $GL_n(\mathbb{F}_p)$ is solvable is a more complex task. I will not provide a precise proof but rather present the ideas and references to reach such a proof.
Let us start with the determinant morphism. We have a morphism $GL_n(\mathbb{F}_p) \longrightarrow \mathbb{F}_p^*$. The multiplicative group of a finite field is a cyclic group of order $p-1$ (see Wikipedia page on finite fields) and thus is solvable. The kernel of the determinant map (morphism) is denoted $SL_n(\mathbb{F}_p)$. We thus have $Aut(G)$ is solvable iff $SL_n(\mathbb{F}_p)$ is.
Now, consider the subgroup of scalar matrices (all entries zero except on diagonal) with determinant one. This group is also solvable (as a subgroup of the group of all diagonal matrices, which is isomorphic to $C_{p-1}^n$). So we can conclude that $Aut(G)$ is solvable iff the quotient, noted $PSL_n(\mathbb{F}_p)$ is.
Last, it is a classical complex but well-known result that $PSL_n(\mathbb{F}_p)$ is simple non abelian except when $n=1$ and also when $n=2$ and $p \in \{2,3\}$. A reference, among many others, is Robert A. Wilson / The Finite Simple Groups / Graduate Texts in Mathematics / Springer.
To conclude, $Aut(G)$ is solvable only when $n$ is 1for all values of $p$ and when $n$ is 2 and $p$ is 2 or 3.