$G = HN$ a group, $N$ normal and $N \cap H =1$. If the conjugacy of $H$ on $N$ has a orbit with all of the non trivial elements: $|N|$ is a prime

Question

Let $G$ be a finite group, $H, N \leqslant G $ with $N$ normal in $G$ such that $G=HN$ and $N \cap H =\{1\}$. Suppose that all of the non-trivial elements of $N$ are in a single orbit for the conjugacy action of $H$ on $N$, then: exists a prime $p$ such that $\forall x \in N$, $x \neq 1$ we have $|x|=p$.

So I can deduce that the conjugacy action of $H$ on $N$ has two orbits: $\{1\}$ and $N \setminus \{1\}$ and if $x \in N \setminus \{1\}$, for the orbit-stabilizer theorem we have that $|N|-1 = [H:H_x]$, so $|N|-1$ divides the order of $H$, hence $|G| = |N|(|N|-1)k$. Now, to show that $|x| = p\,\,$, I can imagine that the strategy would be to show that $|N|$ is the prime $p$. Any hint will be appreciated since I am stuck here, thank you.

Answer 1

The subject line statement is false; it is different from the problem in the body of the post.

As Derek Holt noted, since every nontrivial element of $N$ has the same order (they are conjugate), if $|N|\neq 1$ then there is an element of prime order, and therefore every nontrivial element of $N$ has order that prime $p$. If $|N|=1$, then the statement holds vacuously.

However, this does not mean that $N$ itself is of prime order. That claim is false. For one thing, as phrased in the subject line, $N=\{e\}$ will provide a counterexample by vacuity. But even excluding the case of trivial $N$, the statement is still false.

A counterexample is provided by the holomorph of the Klein $4$-group: let $K=C_2\times C_2$ be the Klein $4$-group, and let $S_3$ act on $K$ by permuting the nontrivial elements. Let $G=K\rtimes S_3$. Then $G$ satisfies the given hypotheses with $N=K$ and $H=S_3$; but $|N|=4$ is not a prime.

Answer 2

From hypotheses we obtain: $$N=\{1_G\}\cup\{h n_0 h^{-1}|h\in H\},\;\;\text{ for any } n_0\in N\backslash{\{1_G\}}.$$ Let $n\in N\backslash{\{1_G\}},$ then for a proper $h\in H$ : $$n^{|n_0|}=(h n_0 h^{-1})^{|n_0|}=h n_0^{|n_0|}h^{-1}=1_G. $$ $$\implies |n|\bigg{|} {|n_0|}\stackrel{\text{simmetry}}{\implies} {|n_0|}\bigg{|} |n|\implies {|n_0|}= |n|.\;\;\; (1)$$ From the Cauchy's theorem exists $n\in N\backslash{1_G}$ s.t. $|n|=p$, for a proper prime (that divides the order of $|N|$), hence from (1), every non-trivial elements of $N $ has order $p$.