Let $G$ be a finite permutation group acting transitive and non-regular on $\Omega$ with $|\Omega| \ge 4$. Suppose further that every nontrivial element has at most two fixed points.
Now let $\alpha \in \Omega$ and suppose that $p \in \pi(G)$ is such that $O_p(G) \ne 1$ and that $|G_{\alpha}|$ is odd. If $|\alpha^{O_p(G)}| \le 2$, then $p = 2$ and $O_2(G) = Z(G)$. Otherwise $|\alpha^{O_p(G)}| \ge 4$.
Why does the above hold? I see that if $|\alpha^{O_p(G)}| = 2$, then as by orbit-stabilizer $2 = |O_p(G) : (O_p(G))_{\alpha}|$ we must have $p = 2$. If $|\alpha^{O_p(G)}| = 1$ then $O_p(G) \le G_{\alpha}$, which implies, as $|G_{\alpha}|$ is odd and $O_p(G) \ne 1$, that $p > 2$. So I guess this case must somehow excluded, but I have no idea how. Also why $O_2(G) = Z(G)$ I do not see, and why $|\alpha^{O_p(G)}| = 3$ is not possible?