$\renewcommand{\phi}{\varphi}$$\DeclareMathOperator{\Aut}{Aut}$Let $N, H$ be groups, and $\Aut(N)$ the automorphism group of $N$. Let us decide how we can compose on $\Aut(N)$, let us say we do it left-to-right, that is, if $\phi, \psi \in \Aut(N)$, then in the composition $\phi \circ \psi$ it is $\phi$ that acts first. In $\Aut(N)^{op}$ composition will be right-to-left, then.
Consider a homomorphism $f : H \to \Aut(N)$, and construct the semidirect product $H \ltimes N$ with multiplication
$$
(h_{1}, n_{1}) (h_{2}, n_{2})
=
(h_{1} h_{2}, n_{1}^{f(h_{2})} n_{2}).
$$
(I write automorphisms as exponents.)
With the identifications you have provided, in this group we have
$$
h^{-1} n h = n^{f(h)}.
$$
Now consider the map $f'$
given by $f'(h) = f(h^{-1})$. It is easily seen that
$$
f' : H \to \Aut(N)^{op}
$$
is a homomorphism. Moreover, it is easily seen that every homomorphism $H \to \Aut(N)^{op}$ is of the form $f'$, for some homomorphism $f : H \to \Aut(N)$.
Now consider the previous semidirect product, and consider in it, with your identifications,
$$
h n h^{-1}
=
(h, 1) (n, 1) (h^{-1}, 1)
=
(h, n) (h^{-1}, 1)
=
(1, n^{f(h^{-1})})
=
n^{f'(h)}.
$$
So $f, f'$ give the same semidirect product, it is just that you read the action of $f(h)$ as a conjugation $n \mapsto h^{-1} n h$ (that is, a right action, as it corresponds to left-to-right composition in $\Aut(G)$), and that of $f'(h)$ as $n \mapsto h n h^{-1}$ (that is, a left action, as it corresponds to right-to-left composition in $\Aut(G)^{op}$)).
The theorem could perhaps be better stated as follows:
Theorem Let $H$ and $K$ be groups and let $\varphi : H \to \operatorname{Aut}(K)$ be a homomorphism. There is a unique group $G$, up to isomorphism, such that $K$ and $H$ are subgroups of $G$, $G$ is the internal semidirect product of $K$ and $H$, and the conjugation action of $H$ on $K$ is $\varphi$.
This is easy to prove, so you should definitely give it a shot!
Notably, this does not mean that each map $\varphi$ gives a different isomorphism class of group $K \rtimes_\varphi H$! For example, there is only one group of order $21$ (up to isomorphism), but there are $6$ distinct actions $\mathbb{Z}/3\mathbb{Z} \to \operatorname{Aut}(\mathbb{Z}/7\mathbb{Z})$! Thus, each of these six actions constructs an isomorphic group of the form $\mathbb{Z}/7\mathbb{Z} \rtimes \mathbb{Z}/3\mathbb{Z}$.
In other words, if two semidirect products of $K$ and $H$ have the same conjugation action, then they must be isomorphic, but isomorphic semidirect products need not have the same conjugation action.
Best Answer
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 $Z(N)$.