Semi direct product $G \cong N \rtimes_{\varphi} K$

So Im having some hard time in trying to understand the semi direct product. So I get that given two groups $N$ and $K$ and a homomorphism $\varphi:K \to Aut(N)$ we can define the semi direct product $N \rtimes_{\varphi} K$ such that:

$(G,\cdot)$ with $G=\{(n,k):n\in N, k\in K\}$ and $\cdot$ defined by the action of $K$ in $N$:

$$(n,k)\cdot(n,k')=(n \varphi(k)(n'),kk')$$

Is a group. Now $\varphi:K \to Aut(N)$ is such that $\varphi(k):N \to N$ is $\textit{necessairily}$ conjugation? In that case, how does the election of $\varphi$ can modify the semi direct product?

• For each $k$, the automorphism $\varphi(k)$ is a conjugation in $G$ but not a conjugation by an element of $N$ necessarily, i.e. not necessarily an inner automorphism. – Levent May 1 '17 at 17:45

When you are given a group $G$ with normal subgroup $N\lhd G$, $K\leq G$, $NK=G$ and $N\cap K=1$ you automatically get $$G\cong N\rtimes K$$ where the homomorphism $\phi:K\to Aut(N)$ is given by $\phi(k)(x)=kxk^{-1}$ (one simply checks that this definition makes the map $N\rtimes K\to G$, $(n,k)\mapsto nk$ an isomorphism).
However, when $N$ and $K$ are arbitrary groups (i.e., they don't come with some ambient group that contains them) you can look at all possible homormophisms $$\phi:K\to Aut(N).$$ Given any such homomorphism, you can form $N\rtimes_\phi K$. For example, there is always the trivial homomorphism $\phi(k)=1_N$ for all $k\in K$. In this case we get back $$N\rtimes_\phi K=N\times K.$$ Sometimes, there are nontrivial maps. For example, there is an isomorphism $\mathbb{Z}_2\cong Aut(\mathbb{Z}_3)$. This automorphism can be realized by sending the generator $1\in \mathbb{Z}_2$ to the map $\varphi_2:\mathbb{Z}_3\to\mathbb{Z}_3$, $\varphi_2(x)=2x$. This gives rise to the group $\mathbb{Z}_3\rtimes\mathbb{Z}_2$ with multiplication \begin{align} (x,\epsilon)(y,\varepsilon)&=(x+2^\epsilon y,\epsilon+\varepsilon)\\ \end{align} for $x,y\in\mathbb{Z}_3$ and $\epsilon,\varepsilon\in\mathbb{Z}_2$.
This last group is isomorphic to $S_3\cong\langle(123)\rangle\rtimes\langle(12)\rangle$. If we set $a=(123)$ and $b=(12)$, then every element can be written as $a^xb^\epsilon$ for some $x\in\mathbb{Z}_3$ and $\epsilon\in \mathbb{Z}_2$ and multiplication is given by $$a^xb^\epsilon a^yb^\varepsilon=a^x(b^\epsilon a^y b^{-\epsilon})b^\epsilon b^\varepsilon=a^{x+2^\epsilon y}b^{\epsilon+\epsilon'}$$
A useful exercise is to look at all homomorphisms $\phi:\mathbb{Z}_4\to Aut(\mathbb{Z}_5)\cong\mathbb{Z}_4$ and construct the associated semi-direct products.