I find it helpful to distinguish between the inner semidirect product and the outer semidirect product. (Similiar to the distinction between the inner direct sum and the outer direct sum of vector (sub)spaces.)
Given a group $G$ and two subgroups $H$ and $N$ of $G$, the group $G$ is called the inner semidirect product of $H$ and $N$ with $N$ normal, if we have $NH = G$ and $H \cap N = 1$, and $N$ is normal in $G$. We then write
$$
G = N \rtimes H \,.
$$
(This notion of an inner semidirect product does not depend on any homomorphism.)
Given on the other hand any two groups $N$ and $H$ and a group homomorphism $\theta \colon H \to \operatorname{Aut}(N)$, the corresponding outer semidirect product $N \rtimes_\theta H$ is defined as the set $N \times H$ together with the multiplication
$$
(n_1, h_1) \cdot (n_2, h_2) := (n_1 \theta(h_1)(n_2), h_1 h_2) \,.
$$
Note that this group $N \rtimes_\theta H$ very much depends on $\theta$.
Now what is the connection between these two?
If $G$ is a group and $H$ and $N$ are two subgroups of $G$ such that $G = N \rtimes H$, then $N$ is normal in $G$. The conjugation action of $H$ on $G$ does therefore restrict to an action of $H$ on $N$. From this we get a group homomorphism
$$
\theta \colon H \to \mathrm{Aut}(N)
\quad\text{given by}\quad
\theta(h)(n) = hnh^{-1} \,.
$$
It can then be checked that the map
$$
N \rtimes_\theta H \to G = N \rtimes H, \quad (n,h) \mapsto n \cdot h
$$
is a group isomorphism. So every inner semidirect product can be seen as an outer semidirect product via the conjugation action of $H$ on $N$.
If on the other hand $N$ and $H$ are two groups and $\theta \colon H \to \operatorname{Aut}(N)$ a group homomorphism, then we can regard both $N$ and $H$ as subgroups $\widehat{N}$ and $\widehat{H}$ of the outer semidirect product $N \rtimes_\theta H$ via the inclusions
\begin{alignat*}{2}
N &\to N \rtimes_\theta H \,, &\quad n &\mapsto (n,1) \,,
\\
H &\to N \rtimes_\theta H \,, &\quad h &\mapsto (1,h) \,.
\end{alignat*}
It then follows for those subgroup $\widehat{N}$ and $\widehat{H}$ that both $\widehat{N} \cdot \widehat{H} = \widehat{N} \rtimes_\theta \widehat{H}$ and $\widehat{N} \cap \widehat{H} = 1$, and that the subgroup $\widehat{N}$ is normal in $N \rtimes_\theta H$. The group $N \rtimes_\theta H$ is therefore the inner semidirect product of its subgroups $\widehat{N}$ and $\widehat{H}$, with $\widehat{N}$ normal.
PS: So given any two groups $H$ and $N$ there may exist multiple outer semidirect products $N \rtimes_\theta H$ because we can choose different group homomorphisms $\theta \colon H \to \operatorname{Aut}(N)$. But if we are given an inner semidirect product $G = N \rtimes H$ (i.e. we are given the group $G$ and its subgroups $H$ and $N$ such that $G = N \rtimes H$) then we are also implicitely given a group homomorphism $\theta \colon H \to \operatorname{Aut}(N)$ with $G \cong N \rtimes_\theta H$: The homomorphism $\theta$ is hidden in the group structure of $G$ and can be retrieved via the conjugation action of $H$ on $N$.
Let $H=S_3$ and $K=\langle t \rangle\cong S_2$.
Let $\phi:t\mapsto i_{(12)}$, the inner automorphism induced by $(12)$.
Form the semi-direct product $G=H\rtimes_{\phi} K$.
Then $G\cong S_3\times S_2$, but the "natural" $K$ in $G$ is not normal; the normal subgroup of order $2$ is generated by $((12),t)$.
Best Answer
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.