# The kernel of an action on the orbits of normal subgroup if group acts such that $\mbox{fix}(g) \in \{0,p\}$ for $g \ne 1$

Let $G$ be a permutation group acting transitively on $\Omega$ and suppose $N \unlhd G$ is a normal subgroup of $G$. Assume that for $g \in N_g(G_{\alpha})$ we have $$G_{\alpha} \cap G_{\alpha}^g = 1$$ for some point stabilizer $G_{\alpha}$. Also assume that every nontrivial element either has no fixed point or exactly $p$ fixed point for some prime $p$. Also suppose $N$ has order coprime to $p$.

The orbits of $N$ form a system of blocks, and we have a natural action of $G$ on these blocks. Under the stated assumptions, do we have that the kernel of this action on the orbits of $N$ is precisely $N$, i.e. $$\forall \alpha \in \Omega: (\alpha^N)^g = \alpha^N \quad \Leftrightarrow \quad g \in N.$$ Does this hold and how to proof it? In general this is not true, as could be seen here.

The dihedral group of order $12$ in its natural action on $6$ points, $$G = \langle (1,2,3,4,5,6), (1,4)(2,3)(5,6) \rangle$$ with $p=2$ is a counterexample.
We can take $N = \langle (1, 5, 3)(2, 6, 4) \rangle$ of order $3$. The kernel of the action on the orbits of $N$ is the subgroup $\langle N,(2, 6)(3, 5) \rangle$ of order $6$.
More generally, for any distinct primes $p,q$ with $q>2$, the group $D_{2q} \times C_p$ has a transitive permutation representation of degree $pq$ and a normal subgroup $N$ of order $q$ such that the kernel of the action on the orbits of $N$ is $D_{2q}$.