Let $G$ be a finite transitive permutation group on a set $\Omega$ of odd degree. Suppose that the four-point stabilizers are trivial and that some three-point stabilizer is nontrivial; this is equivalent to the condition that every nontrivial element has at most three fixed points, and some nontrivial element actually fixes three points.
As $|G : G_{\alpha}| = |\Omega|$ is odd, $G_{\alpha}$ contains a Sylow $2$-subgroup of $G$. Let $S$ be such a Sylow $2$-subgroup in $G_{\alpha}$. Suppose that $S$ fixes three points, denote them by $\Delta := \{\alpha,\beta,\gamma\}$. Let $M_0 = G_{\alpha}\cap G_{\beta}\cap G_{\gamma}$ be the point-wise stabilizer of $\Delta$ and $M := N_G(M_0)$.
I want to show that $N_G(M) \le M$.
As $M = N_G(M_0)$, the group $M$ acts on $\Delta$. Hence $|N_G(M_0) : N_G(M_0) \cap G_{\alpha}| \le 3$. Likewise $N_G(M)$ acts on the orbits of $M$, and $M$ has at most $3$ orbits of size $1$. If $N_G(M)$ acts on $\Delta$ (i.e. permutes the fixed points of $M_0$), then it normalizes $M_0$. So I guess we must somehow show either $\Delta$ is the unique orbit of $M$ of length $3$, or $M$ also fixes $\Delta$ point-wise.
We have $S \le M_0 \le M \le N_G(M)$ and $S \le M_0$ gives that the orbits of $N_G(M)$ and of $M$ on $\alpha, \beta, \gamma$ must have odd size. If we look at the orbits of $M_0$, then it has three orbits of size $1$ (its fixed points) and the other orbits must have length $|M_0|$, which is even. Maybe from this it might be possible to conclude that $M$ has a unique orbit of size three.
So do you have any ideas how to proceed?