As written on Wikipedia:Frobenius_group
The Frobenius kernel $K$ is uniquely determined by $G$ as it is the Fitting subgroup, and the Frobenius complement is uniquely determined up to conjugacy by the Schur-Zassenhaus theorem. In particular a finite group $G $is a Frobenius group in at most one way.
On the other side, in Kurzweil/Stellmacher, in chapter 4 about Frobenius groups I read:
Let $G$ be a Frobenious group with complement $H$ and kernel $K$. Let $H_0$ be another Frobenius complement of $G$ such that $|H_0| \le |H|$. Then $H_0$ is conjugate to a subgroup of $H$.
But with regard to the uniqueness as written on wikipedia we would even have that $H_0$ is conjugate to the whole of $H$. Right? And this is a much stronger. I am wondering why they state such a weaker result. This book is a standard reference and the uniqueness is used in some arguments, for example in this post of mine where it is essential that the different point stabilizers on the different orbits are conjugate in $N$, i.e. $N_{\alpha} = N_{\beta}^n$ if $\alpha$ and $\beta$ are in different orbits with point stabilizer $N_{\alpha}, N_{\beta}$ on each one. Hence $N_{\alpha}$ fixes $\beta^n$ in the other orbit.
Am I right in my objection? And why then they choose to present a much weaker result?