Let $G < S_\Omega$ be a group of permutations of a finite set $\Omega$ such that it is transitive, and has at least one, non trivial, system of imprimitivity $\mathcal{P} = \{B_1, \dots, B_m\}$.
I have the following conjecture: for any pair of blocks $B_i$ and $B_j$ the intersection of their stabilizers is not trivial, i.e., $$\{id\} < Stab_G(B_i) \cap Stab_G(B_j)$$ Where $id$ is the identity permutation.
By the moment my approach to this is the following. Consider $h \in Stab_G(B_i)$. For $x \in \Omega$ denote the minimal whole number $k\geq 0$ such that $x\cdot h^k = x$ as $o_h(x)$. As well for $B_j \in \mathcal{P}$, $o_h(B_j)$ the minimal whole number such that $B_j \cdot h^k = B_j$. For example $o_h(B_i) = 0$. We can see that if $x \in B_j$ then $$o_h(x)\geq o_h(B_j)$$ the problem would be if for every $h \in Stab_G(B_i)$ we have that $o_h(x) = o_h(B_j)$. By the moment I have not found neither a counter example to the conjecture nor why the inequality above is nor strict for every permutation in the stabilizer.
Any hint or reference for this would be appreciated.
This does not hold. Small counterexample:
$$ g:=\langle (1,3,5)(2,4,6), (1,4)(2,3)(5,6) \rangle\cong S_3 $$ Then $\{\{1,4\},\{2,5\},\{3,6\}\}$ is a block system, but the stabilizers of $\{1,4\}$ and $\{2,5\}$ intersect trivially.
This is not dependent on the representation being regular. For example take $$ g=\langle (1,3,5,8,10,12,14,15)(2,4,6,7,9,11,13,16), (1,5,16)(2,6,15)(7,10,14)(8,9,13)\rangle\cong GL(2,3)$$ and the blocks $\{1,2\}$, $\{3,4\}$ (in the same block system).
But what it of course requires is that the permutation representation is not ``minimal'' -- if the action on the blocks has a nontrivial kernel, this kernel will lie in every block stabilizer.