The following is part of an exercise from Permutation Groups by Dixon and Mortimer (Exercise 2.7.1); Show that a primitive group $G$ of degree $> 1$ is not regular if a point stabilizer equals its normalizer $ N_{G} (G_{ \alpha} ) $.
My attempt so far is as follows. Suppose $G_{ \alpha } \neq N_{G} (G_{ \alpha } ) $ for any $ \alpha $. As $G$ is primitive, each point stabilizer is a maximal subgroup of $G$.
Now $ G_{ \alpha } < N_{G} (G_{ \alpha } ) \leq G$ and so $N_{G} (G_{ \alpha } ) = G $ by maximality of $G_{ \alpha } $. Thus $ G_{ \alpha } $ is a normal subgroup of $G$. Since all point stabilizers of a transitive group action are conjugate, $ G_{ \alpha } $ must stabilize every point.
I am not sure how to proceed here. If the action is faithful, then it must be that $G _{ \alpha } = \{ 1 \} $, which shows the action is regular and we are done.
I'm not sure how to deal with the situation where the action is not faithful. I also think this might just be some missing information from the book itself. Most websites seem to agree that the definition of a "primitive group" requires that the group has a faithful primitive group action on a set. However the textbook does not appear to include "faithful" in the definition of primitive group.