Suppose that a finite group $G$ admits a Frobenius group of
automorphisms $F H$ with kernel $F$ and complement $H$ such that $F$
acts without nontrivial fixed points (that is, such that $C_G(F)=1$).
It is proved by Belyaev and Hartley in [Centralizers of finite nilpotent subgroups in locally finite groups](https://doi.org/10.1007/BF02367023) that $G$ is a solvable group.
So, there are some papers studied on the Fitting height of $G$, exponent of $G$, rank of $G$.
For example, a paper [Fitting height of a finite group with a Frobenius group of automorphisms](https://doi.org/10.1016/j.jalgebra.2012.05.011) by E.I. Khukhro. 
I am curious about the following question.

**Question: If $G$ is an abelian group which is $FH$-indecomposable and $F$ acts fixed-point-freely on $G$, I am wondering whether it is true that $G$ is homocyclic $p$-group.**

Obviously, $G$ is a $p$-group for some prime $p$.
If the action of $FH$ on $G$ is coprime, then a result of M. Harris shows that $G$ is homocyclic without assuming that $FH$ being a Frobenius group.
So, we may assume that $p\mid |H|$.
If $G$ is an elementary abelian group, then $G$ has a basis which is permuted by $H$ by Theorem 15.16 in "[Character theory of finite groups](https://doi.org/10.1090/chel/359)" by I.M. Isaacs.

Here is an example of a very special case: $G$ is an abelian $2$-group which is $FH$-indecomposable and $FH\cong S_3$.
Then $$G=C_G(H)\times C_G(H)^x$$ where $F=\langle x\rangle$.
Since $F$ acts fixed-point-freely on $G$, $C_G(H)$ is a cyclic subgroup.
So, $G$ is a homocyclic 2-group. Also, power map defines isomorphism of $FH$-chief factors of $G$.
 
Any explanation, references, suggestion and examples are appreciated.