Let $G$ be a nonregular, transitive permutation group on $\Omega$ such that each nontrivial element has at most two fixed points. Suppose $S \in \mbox{Syl}_2(G)$ and that we have $\alpha, \beta \in \Omega$ distinct such that $S_{\alpha} = S_{\beta}$ has index $2$ in $S$ and every element in $S \setminus S_{\alpha}$ interchanges $\alpha$ and $\beta$ (this follows as $S$ normalizes $S_{\alpha}$). Now assume $\{\alpha,\beta\}$ is the unique $S$-orbit of length two, and all other $S$-orbits have length $|S|$.
Now take some $x \in S \setminus S_{\alpha}$ and look what it does on the $S$-orbits on $\Omega \setminus \{\alpha,\beta\}$.
Why does every orbit of $X := \langle x \rangle$ on $\Omega \setminus \{\alpha, \beta\}$ has the same size?
These orbits of $X$ (or $x$ for short) are cycles if we view $x$ as an element of the symmetric group on $\Omega$.
There must be one very short and easy way to see it; but I do not see it???
Suppose some $x^k$ has a fixed point, then as $x^{2k} \in S_{\alpha} = S_{\beta}$ (as $S / S_{\alpha}$ has order two) has this fixed point too by hypothesis we must have $x^{2k} = 1$. This shows $|X_{\omega}| \le 2$, and if $o(x) = 2^m$, then only $x^{2^{m-1}}$ is a candidate to fix some point. This is all I got, so the orbits have either size $|X|$ or $|X| / 2$, but I do not get that they all have equal size?