Transitivity of the action of a normalizer on the set of fixed points

Question

Let $G$ be a finite group acting transitively on a set $X$ (from the left). Let $H$ be a subgroup. Denote by $S$ the set of fixed points of $H$. That is $$S=\{x\in X\mid \text{$h\cdot x=x$ for all $h\in H$}\}.$$

Let $N=N_G(H)$ be the normalizer. If $x\in S$ and $n\in N$, then $H\le Stab_G(x)$. Hence $H=nHn^{-1}\le Stab_G(n\cdot x)$, allowing us to conclude that $n\cdot x$ is also a fixed point of $H$. Therefore $N$ acts on the set $S$, and we have reached my question:

Is the action of $N$ on $S$ necessarily transitive?

Testing with small groups suggests this to be the case, but I am nowhere near seeing why this should always happen. It might also be false!

This question is my translation of a natural question from Galois theory into the language of group actions. Incarnations: 1, 2.

An answer by Derek Holt shows that the answer is affirmative, if $H$ happens to be a Sylow subgroup of $G$. What happens in general?

Answer 1

For a counterexample, how about $S_4$ acting on the set of six pairs of points $\{ \{a,b\} : a, b \in \{1,2,3,4\}, a \ne b\}$.

The image of $(1,2)$ under this action has fixed set $\{\{1,2\},\{3,4\}\}$. The normalizer of the subgroup $H$ of order 2 that it generates is the image of the subgroup $\langle (1,2), (3,4) \rangle$, which fixes both of those points. So it does not act transtiviely on the set of fixed points.