Classifying $G$ with $|G|=pq$ has been asked several times in this site, but I am having trouble with number theory. Let $q<p$, $Q=\langle u\rangle\in \text{Syl}_q(G)$, $P=\langle v\rangle\in \text{Syl}_p(G)$. We know with a little work using the Sylow Theorems that that $P\lhd G$ and $P\cap Q=1$, so $G=P\rtimes Q $ as one can see in this previous Math StackExchange post. So we need to study possibilities for $\theta: Q\rightarrow \text{Aut}(P)$.
It sufficies to study what $\theta(u) $ does, because that would teach us that $\theta(u^j)=\theta(u)\circ...\circ \theta(u)$. By a similar reasoning we only need to study $\theta(u)(v)$. Let us do this:
$$\theta(u)(v)=v^j\Rightarrow$$
$$v=I(v)=\theta(u^q)(v)=(((v^j)^j)^{...})^j=v^{j^q}$$
This in turn means $|v|=p|j^q-1 $. So we need to study values of $j$ for which $j^q\equiv 1\mod p$. If $j=1$, this means $\theta(u)=I$ and $G\approx \mathbb{Z}_q\times \mathbb{Z}_p$. So my question boils down to:
If $q<p$ (where $q,p$ are prime numbers), find all $j\in \mathbb{N}$ with $ p > j>1 $ such that $j^q\equiv 1 \mod p $.
The order of $j$ must divide $q$ and also $p-1$ by Fermat's Theorem $(j^{p-1}\equiv 1 \mod p)$ and it is not equal to $1$ because of the range of $j$ values we are considering. So the order of $j$ must be $q$ and $q|p-1$ by Fermat's Theorem.
Note that this teaches us that $v^{j^l}$ is not equal to $v$ when $l=1,2,...,q-1$, and it returns to $v$ only when $l=q$. This means $\theta(u^l)$ is always different. In other words, $\theta$ is injective! (We will need this later).
I think there isn't a self contained argument in MSE, so I decided to write it down for the sake of completeness. Let us go back to our congruence. Using Anne's comment, we are interested in finding elements in $\mathbb{Z}_p^*$ whose order is $q$. These are in the unique cyclic subgroup of order $q$ in $\mathbb{Z}_p^*$. Thus there are $\phi(q)=q-1$ generators. So there are $q-1$ values satisfying my congruence in $j$.
At first sight this might indicate there are several possibilities for $P\rtimes Q$, but that is not the case and we will prove this happens essentially because there is only one cyclic subgroup of order $q$ in $\mathbb{Z}_p^*$. That means that if $j_1$ (defining $\theta_1$) and $j_2$ (defining $\theta_2$) satisfy our criterium, we must have $j_2=j_1^{z_1}$:
$$\theta_2(u)(v)=v^{j_2}=v^{j_1^{z_1}}=\theta_1(u^{z_1})(v).$$
Let us define $\Psi: P\rtimes_{\theta_2}Q\rightarrow P\rtimes_{\theta_1}Q$ by:
$$\Psi(x,y)=(x,y^{z_1})$$
$\Psi\left((x,y)(z,w)\right)=\Psi(x\theta_{2}(y)(z),yw)=(x\theta_1(y^{z_1})(z),(yw)^{z_1})$. Because $Q$ is cyclic (hence abelian) we can simplify this further and make sure it is a homomorphism: $\Psi\left((x,y)(z,w)\right)=(x\theta_1(y^{z_1})(z),y^{z_1}w^{z_1})=(x,y^{z_1})(z,w^{z_1})=\Psi(x,y)\Psi(z,w)$
Analogously, we can write $j_1=j_2^{z_2}$ and this defines a new homomorphism $\tilde{\Psi}:P\rtimes_{\theta_1}Q\rightarrow P\rtimes_{\theta_2}Q$ with $\tilde{\Psi}(x,y)=(x,y^{z_2})$. This we will show is an inverse to $\Psi$. We only need to show it is a right inverse because $|P\rtimes Q|=qp$. If we compose $\Psi \tilde{\Psi} (x,y)=(x,y^{z_1z_2})$. We have $\theta_2(u)=\theta_1(u^{z_1})=\theta_1(u)^{z_1}=\theta_2(u^{z_2})^{z_1}=\theta_2(u^{z_1z_2})$. So $u^{z_1z_2}=u$ ($\theta_2$ is injective!). In particular, $\Psi \tilde{\Psi} (x,y)=(x,y^{z_1z_2})=(x,y)$. So we really have an isomorphism!
This appears to be correct and proves there is a unique non abelian group, but it is very ugly. I am on the lookout for a nicer answer. Sorry about constantly changing what I want from the post.
**text**for text. – Shaun May 06 '23 at 23:07