I have a two parameter family of moebius transform $A_{q,p}$ with $q,p\in\mathbb{N}$ and $$A_{q,p}=\begin{pmatrix}0& 2q \\ p&q\end{pmatrix}$$
Additionally, I assume that $A_{q,p}(1)=1$ for all $p,q$. Using the product of the two fixed points, which gives $\dfrac{2q}{p}$, I deduced that necessarily $p=q$ by the formula for the two fixed points I found here on wikipedia. But this would mean that the whole family only depends on one parameter and its action on the upper half plane is equivalent to $$A_{q,p}=\begin{pmatrix}0& 2 \\ 1 &1 \end{pmatrix}.$$
This seems somehow counterintuitive. Is this reasoning correct or did I fail to see something here?