Uniform subdegrees force an imprimitive action to be Frobenius

For a transitive finite permutation group, suppose each point stabilizer has orbits of a common size $r$ on the other points. If a nontrivial block system exists, all two-point stabilizers are trivial. Indeed for block size $b$, $r\mid n-1$ gives $\gcd (r,b)=1$. An orbit of a block not containing $\alpha$ has size at most $br$ and divisible by $br$, so its b point-orbits are disjoint. A stabilizer of $\alpha$ and a point of that block fixes the whole block. Interchanging the two points also fixes the block of $\alpha$. Equal two-point-stabilizer orders then show that each two-point stabilizer fixes every point, and faithfulness makes it trivial.