How many non-abelian groups of order $lpq$ are there? 
If $l,p,q$ are primes with $l<p<q$, such that
$$p\nmid (q-1)\hspace{1cm} l\mid (p-1)\hspace{1cm} l\mid (q-1) $$
I want to show that there are at least $1$ and at most $(l+1)$
  non-abelian groups of order $lpq$.

I already showed (using Sylow theorems) that in this case, if $G$ is a group with order $lpq$ then 
$$ G \cong C_{pq}\rtimes C_l .$$
The fact that there is at least one non-abelian group I showed the following way: we know that $|\text{Aut}(C_{pq})| = (p-1)(q-1)$, so $l$ divides its order, and as a consequence of the Sylow theorems we know that $C_l\hookrightarrow \text{Aut}(C_{pq})$, thus there is a non-trivial homomorphism which gives us a non-abelian group $C_{pq}\rtimes C_l$. 
The problem I'm having is showing that $\text{Aut}(C_{pq})$ can have at most $(l+1)$ copies of $C_l$ in it. Any hints?

An answer to your question is provided here (page 73):
http://users.hawknetwork.org/~adam/ClassifyPQR-FINAL.pdf
You'll see that if $l$, $p$, $q$ satisfy your conditions then there are always $p+1$ non-abelian groups of order $lpq$.
Another reference for that fact is the book by Blackburn, Neumann, and Vekataraman "Enumeration of finite groups", page 238.

