$G = HN$ 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 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.

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 $p$.
