Let $G$ be a finite transitive permutation group of degree $n$, the list of subdegrees (the lengths of the orbits of one of the point stabilizers) is an invariant of $G$. We shall denote them in increasing order: $n_1 = 1, n_2, \ldots, n_r$ where $r$ is the rank of $G$.
Show that if either $n_i \ge n/2$ or $n_i \ge n/4$ and the corresponding orbital $\Delta_i$ is not self-paired, then any two vertices in the orbital graph of $\Delta_i$ can be joined by a nondirected path of length at most $2$.
This is taken almost verbally from Dixon & Mortimer, Permutation groups, page 72. An orbital is an orbit of $G$ in its induced action on $\Omega\times\Omega$. There exists a bijection between the orbitals and the orbits of some (fixed) point stabilizer (in this way we can speak about corresponding orbitals as in the question), and the orbits for different point stabilizers could also be mapped onto each other. I can give details if wanted, but they could also be found in the above book on page 66. If we see an orbital as a relation over $\Omega$, we can look at its inverse relation (just swap the tupels), and we call an orbital self-paired if it equals its inverse. Also an orbital, as a subset of $\Omega\times\Omega$, could also be interpreted as a graph, its orbital graph, which is interpreted as an undirected graph, i.e. an element $(\alpha, \beta) \in \Delta$ works as an edge in either direction.
For an orbital $\Delta$ and $\alpha \in \Omega$ set $\Delta(\alpha) = \{\beta\in\Omega : (\alpha, \beta) \in \Omega \}$, hence $$ \Delta = \bigcup_{\alpha \in \Omega} \{\alpha\}\times \Delta(\alpha) $$ and as the union is disjoint we have $|\Delta| = |\Delta(\alpha)||\Omega|$ for some $\alpha\in \Omega$ (where that $|\Delta(\alpha)| = |\Delta(\beta)|$ follows as there exists a bijection).
Now suppose $n_i \ge n/2$ and let $\alpha, \beta \in \Omega$ be arbitrary, suppose $\beta\notin \Delta(\alpha)$ (otherwise there is nothing to show), as $\beta \notin \Delta(\beta)$ (or the orbital would be diagonal, which implies $n_i = 1$) we could distribute $n-1$ points on two sets $\Delta(\alpha)$ and $\Delta(\beta)$ each holding at least $n/2$ elements, hence by a pigeonhole like reasoning they must have at least one point in common, i.e. $$ \Delta(\alpha) \cap \Delta(\beta) \ne \emptyset $$ which shows there exists $\tau \in \Omega$ with $(\alpha,\tau), (\beta, \tau) \in \Delta$, i.e. a path of length at most $2$.
But now if $n_i \ge n/4$ and $\Delta_i$ is not self-paired, I have no idea how to proceed. I would like to apply a similar counting argument as above, but for this I have to show that we have a set $\Gamma \subset \Omega$ of size at most $n/2-1$ such that (supposing again $\beta\notin\Delta(\alpha)$) $$ \Delta(\alpha) \subset \Gamma, \quad \Delta(\beta) \subset \Gamma $$ but here I have no idea how to enforce this, so any hints?