The simplest thing is to remember that a semidirect product $N\rtimes K$ of groups $N$ and $K$ is equivalent to a group homomorphism $\phi\colon K\to\mathrm{Aut}(N)$: given a semidirect product $N\rtimes K$, we obtain $\phi$ by letting $\phi(k)$ be the automorphism induced on $N$ by conjugation by $k$ in $N\rtimes K$. Conversely, given a homomorphism $\phi\colon K\to\mathrm{Aut}(N)$, we construct the semidirect product $N\rtimes_{\phi}K$ as the group whose underlying set is the set $N\times K$ of all ordered pairs $(n,k)$ with $n\in N$, $k\in K$, and with group operation given by $(n,k)\cdot (m,\ell) = (n\phi(k)(m), k\ell).$ In this group, we have that $(e,k)(n,e)(e,k)^{-1} = (e,k)(n,e)(e,k^{-1}) = (e\phi(k)(n)e,kk^{-1}) = (\phi(k)(n),e).$ That is, conjugation by $(e,k)$ corresponds exactly to the application of $\phi(k)$ to $N$. We then identify $N$ with the subgroup $\{(n,e)\mid n\in N\}$, and $K$ with the subgroup $\{(e,k)\mid k\in K\}$.
So to get a semidirect product you want to find some group $K$ and a group homomorphism $\phi\colon K\to\mathrm{Aut}(N)$. Then conjugation by $k$ in $N\rtimes K$ will correspond to the automorphism $\phi(k)$. Since we want every automorphism to be represented, we want $\phi$ to be onto. And that's really all we need.
So... we just need to find some group $K$ that has an onto homomorphism to $\mathrm{Aut}(N)$. What's the simplest one and simplest $\phi$ we can find?