So this is a past paper question that I am currently struggling with:
Suppose $H \cong C_2$ and $K \cong C_3$, and $\phi: H \rightarrow Aut(K)$ is non-trivial. Find an explicit isomorphism between $H \ltimes_{\phi} K$ and $S_3$.
Now, in plain notation, let $H \cong \{1,a \} $ and $K\cong \{1,b,b^2 \}$. Therefore, $Aut(K) = \rho_1, \rho_2 $ and which we are defined by $$ \rho_a = \left( \begin{array}{ccc} 1 & b & b^2 \\ 1 & b & b^2 \\ \end{array} \right), \ \rho_2 = \left( \begin{array}{ccc} 1 & b & b^2 \\ 1 & b^2 & b \\ \end{array} \right).$$ So we let $\phi$ be defined as $ 1 \rightarrow \rho_1$ and $a \rightarrow \rho_2$. Now, I don't know really where to go from here. I think the subset $H \ltimes_{\phi} K $ contains $(1,1), (1,b), (1,b^2), (a,1), (a,b)$ and $(a,b^2)$, but regarding an explicit isomorphism between $S_3$. I am not necessarily sure if I should just provide a mapping. I haven't used anything about the fact we have a semi-direct product and a non-trivial homomorphism.
Isomorphisms preserve order; so if you want an explicit isomorphim between your two groups, you have to send $(1,b)$ to an element of order $3$, such as the cycle $(1,2,3)$, and $(a,1)$ to an element of order $2$, such as $(1,2)$. Now every element of $H\ltimes_{\phi} K$ can be uniquely written as $(a^\alpha,b^\beta)$, with $\alpha=0,1$ and $\beta=0,1,2$; and moreover we can check that $$(a^\alpha,b^{\beta})=(1,b^{\beta})(a^\alpha,1).$$ This suggests to define the isomorphism as $$f((a^\alpha,b^\beta))=(1,2,3)^\beta(1,2)^{\alpha}$$(with the right-hand side product computed in $S_3$), and you can check that this is really an isomorphism.
But it might be easier to do things the other way around. Notice that $S_3$ has a normal subgroup $K=\langle (1,2,3)\rangle$ of order $3$, and a subgroup $H=\langle (1,2)\rangle $ of order $2$. Since $K$ is normal, $H$ acts on $K$ by conjugation; we note $\phi:H\to Aut(K)$ this action, and notice that $$(1,2) (1,2,3)(1,2)=(1,3,2)=(1,2,3)^2.$$ Now one can show that $KH$ is a subgroup (because $K$ is normal) of $S_3$ containing elements of order $2$ and $3$, so it must be equal to $S_3$. Thus every element of $S_3$ can be written as $kh$ with $k\in K$ and $h\in H$; and such an expression has to be unique, because $K\cap H$ is trivial. Now if we compute the product of $kh$ and $k'h'$ in this form, we find $$(kh)(k'h')=k(hk'h^{-1})(hh')=(k\phi_h(k'))(hh').$$ This is exactly the same as a product in $H\ltimes_{\phi} K$, thus we have an isomorphism $kh\mapsto (h,k)$.
This approach amounts to characterize $S_3$ as an inner semi-direct product, and show the link with outer semi-direct product.