Let $G$ be a finite group acting nonregularly and transitive on $\Omega$ such that each nontrivial element has at most two fixed points and $|\Omega| \ge 4$. I know three facts:
i) If $1 \ne X \le G_{\alpha}$, then $|N_G(X) : N_G(X) \cap G_{\alpha}| \le 2$, as $N_G(X)$ acts on the at most two fixed points of $X$ with point stabilizer $N_G(X) \cap G_{\alpha}$.
ii) If $|\Omega| \ge 5$ and $|Z(G)| = 2$. Then $\overline G := G / Z(G)$ acts on the $Z(G)$-orbits and satisfies the above hypothesis, i.e. acts nonregulary, transitive and each nontrivial element has at most two fixed points. Further if the point stabilizers in $\overline G$ have odd order, then $\overline G$ is a Frobenius group.
iii) If $p$ is odd and divides $|G_{\alpha}|$, then $G_{\alpha}$ contains a full $p$-Sylow subgroup of $G$. For if $P \in \mbox{Syl}_p(G_{\alpha})$, then if we look how the $P$-orbits partition $\Omega$ we find that $P$ has at most two orbits of length one, but at least the one $\{\alpha\}$, and all other orbits have size $|P|$ as $P_{\omega} = 1$ for $\omega \notin \{\alpha,\beta\}$. Hence $|\Omega| = |G : G_{\alpha}|$ is not divisble by $p$ and $P$ is a Sylow $p$-subgroup of $G$.
Lemma: Let $\alpha \in \Omega$. Suppose that $p \notin \pi(G)$ is such that $O_p(G) \ne 1$ and that $|G_{\alpha}|$ is odd. Then $G_{\alpha}$ is a Frobenius complement (hence metacyclic) or $p = 2$, $O_p(G) = Z(G)$ and $G / Z(G)$ is a Frobenius group.
Proof: Observe that $O_p(G)G_{\alpha}$ acts nonregulary and transitive on $\Delta := \alpha^{O_p(G)}$. As $O_p(G)$ is a normal, nontrivial subgroup of a faithful group action it could not fix any point, so $|\Delta| > 1$. Also we must have $O_p(G) \cap G_{\alpha} = 1$, in case $p = 2$ this follows as $|G_{\alpha}|$ is odd, in case $p > 3$ this follows by fact iii) as if $O_p(G) \cap G_{\alpha} \ne 1$ then this would imply $O_p(G) \le G_{\alpha}$, which is not possible. If $|\Delta| = 2$ this implies $p = 2$ and $|O_2(G)| = 2$, hence $O_2(G) = Z(G)$. Then fact ii) from above with $$ (G / Z(G))_{\alpha^{Z(G)}} = G_{\alpha} Z(G) / Z(G) \cong G_{\alpha} / (Z(G) \cap G_{\alpha}) \cong G_{\alpha} $$ (as a central element fixing any point would fix every point, hence $Z(G) \cap G_{\alpha} = 1$) implies that $G / Z(G)$ is a Frobenious group. If $|\Delta| = 3$ similarly $p = 3$ and $|O_3(G)| = 3$ and hence $G_{\alpha} \le C_G(O_p(G))$ as $\mbox{Aut}(O_p(G)) = 2$ and $G_{\alpha}$ has odd order, so it must fix every element from $\Delta$, which is not possible as $G_{\alpha} \ne 1$. So $|\Delta| \ge 4$.
So we can suppose $(O_p(G)G_{\alpha}, \Delta)$ acts nonregularly, transitive and each nontrivial element has at most two fixed points with $|\Delta| \ge 4$. If $p$ is odd, then fact iii) above yields that $p$ does not divide $|G_{\alpha}|$. Thus $C_{O_p(G)}(G_{\alpha}) = 1$ by fact i), which means that $O_p(G)G_{\alpha}$ is a Frobenius group with complement $G_{\alpha}$. Hence suppose $p = 2$. If $Z(O_2(G)G_{\alpha}) \ne 1$, then $O_2(G)G_{\alpha}/Z$ is a Frobenius group with complement $G_{\alpha}$ by fact ii). If $Z = 1$, then $G_{\alpha}$ acts fixed point freely on $O_2(G)$ and thus $O_2(G)G_{\alpha}$ is a Frobenius group with complement $G_{\alpha}$. In both cases by a theorem from B. Huppert, Endliche Gruppen, V.8.18, $G_{\alpha}$ is metacyclic. $\square$
Now my questions:
1) "If $Z = 1$, then $G_{\alpha}$ acts fixed point freely on $O_2(G)$"
If we have some $x \in G_{\alpha}$ with $u^x = u$ for all $u \in O_2(G)$, does this imply that $x$ is central? But then it also has to commute with all the other elements from $G_{\alpha}$. But I do not see that?
2) "suppose $p = 2$. If $Z(O_2(G)G_{\alpha}) \ne 1$, then $O_2(G)G_{\alpha}/Z$ is a Frobenius group with complement $G_{\alpha}$ by fact i)"
The point stabilizer in $O_2(G)G_{\alpha}$ is $O_2(G)G_{\alpha} \cap G_{\alpha} = G_{\alpha}$ and $G_{\alpha} Z(G) / Z(G) \cong G_{\alpha}$ is odd in in the factor group, hence the point stabilizers have odd order. But why $|\Omega| \ge 5$?