Conjugation action on a semi-direct product $N\rtimes H$

Question

Let $G$ be a finite group, with $H\le G$ and $N\trianglelefteq G$ such that $G=N\rtimes H$.
Suppose that the conjugation action of $H$ on $N$ induces $2$ orbits in $N$. Prove the following requests:
a) $\exists p$ prime $\forall 1\ne n\in N$ such that $o(n)=p.$
b) SHow that $N$ is abelian.

I said that if $G=N\rtimes H$, hence we have the conditions $(*)\begin{cases}N \cap H=\{1_G\}\\NH=G \end{cases}$. For a generic element, $n \in N$ the stabilizer of $n$ in the action $\varphi:H\to\operatorname{Aut}(N)$ is such that $h^{-1}nh=n\iff h\in C_H(n)$, whose order divides $H$.
Furthermore, if the number of orbits is two, the quotient $H/Ker(\varphi)\cong \varphi(H)\le\operatorname{Aut}(N)$ has order two, with $Ker(\varphi):=\underset{n\in N}\bigcap C_H(n)$. I'm confusing things because the map induced by the semi-direct product has the same form of the map induced by the conjugation action and I'm not sure about how using the Hypothesis $(*)$.
If $N$ is abelian it means that its centralizer is the entire $G$ (with $C_G(H)\le Z(G)$), and in particular, considering tha action restricted to $H$, $C_H(N)$ should corespond to $H$.

Thank for the help and for your patience.

Answer 1

Under conjugation one orbit is just the identity. So all other elements of $N$ are conjugate in $G$ and therefore have the same order.

Let this order be a multiple of the prime $p$, then the $p$th power of an element of $N$ is still in $N$ but has a smaller order i.e. it has to be the identity.

Since $N$ is a $p$-group, $Z(N)$ is non-trivial and is a characteristic subgroup of $N$. Therefore $Z(N)$ is fixed by $H$ and so all elements of $N$ are in $Z(N)$.