For a Frobenius group its kernel is a characteristic and nilpotent group, the last property restricts the possibilities how a given group could be represented as a Frobenius group. A statement of this I found here in Passman's book. I guess this means that the complement $H \le G$ is unique up to conjugation.
Here is an argument were this is used (or I guess it has something to do with the facts I collected above), but I am unsure how it enters into the argument:
Suppose we have a group $G$ acting on some set $\Omega$ such that each nontrivial element fixes no point or exactly $p$ points for some prime $p$, and suppose we have a normal subgroup $N$. Now suppose $G_{\alpha}N$ is a Frobenius group with complement $G_{\alpha}$ acting transitively on each orbit of $N$ (i.e. it has the same orbits). Further suppose $N$ acts also as a Frobenius group on all its orbits. By the uniqueness of the Frobenius representation if follows that $N$ has exactly $p$ orbits.
I would be glad if someone could clarify this and what exactly is meant by "uniqueness of Frobenius representation" and how it enters the argument here?