While reading some group theory notes I came up to this fact:
Proposition: If $G$ is the inner semi direct product of $H,K$ ($G=HK$, $H\cap K=\left\{1\right\}$ and $H\unrhd G$) then $G\cong H\rtimes_fK$ where $f$ is expicitely given by $k\mapsto f_k(h)=h^{-1}kh$.
Now I had no problem with this until I read the Classification of finite groups with order $pq$ for prime $p,q$ where $p\equiv 1\mod q$.
After application of the Sylow Theorems one obtains such $H,K$. But then, the writer turned to the problem of counting the homomorphisms in $f:K\to Aut(H)$. This seems the way all other group theory books deal with this but it is never explained why this is needed.
Unless I am confused, the Proposition implies that $G$, up to isomorphism is determined by $H,K,f$ and $f$ is explicitely given. If $H\rtimes_gK\not\cong H\rtimes_fK$ then it can't be that $G\cong H\rtimes_gK$ as that would contradict the transitivity of $\cong$. If we only had the existence of $f$ in Proposition, then I agree we would need to determine all such $f$, as $G$ would be the product of $H,K$ and one of these $f$. So why do we need to do this even when we have a specific $f$?
$\newcommand{\Aut}[0]{\mathrm{Aut}}$In case this is what you are interested in, given $H$ and $K$, two different $f$ may well lead to isomorphic groups $G$.
For instance, in the case you mentioned, consider the primes $p = 7$ and $q = 3$. Let $K = \langle b \rangle$ and $H = \langle a \rangle$. Consider the homomorphisms $f, g : K \to \Aut(H)$ determined by $$ f : b \mapsto (a \mapsto a^{2}) $$ and $$ f : b \mapsto (a \mapsto a^{4}). $$ Then $$ H \rtimes_{f} K \cong H \rtimes_{g} K. $$
Actually, when $p \equiv 1 \pmod{q}$, you should have seen that there are two isomorphism classes of groups of order $pq$. One, the cyclic group, corresponds to the trivial homomorphism $K \to \Aut(H)$, the other to all other homomorphisms.