Note that $Mr$ denotes a Mobius-transformation i.e. $Mr=\frac{ar+b}{cr+d}$. I want to show that there are infinitely many Matrices $M \in SL_2(\mathbb{Z})$ such that $Mr=s$. Since I had no clue where to start I thought about writing the equation into a linear system.
I wrote $r=\frac{e}{f},s=\frac{g}{h}$ and I think without the loss of generality it is possible to assume $gcd(e,f)=gcd(g,h)=1$ if this is even important. Now I am left with $$ae+bf=g, \quad ce+df=h, \quad ad-bc=1.$$ and don't know how to go on. Furthermore, I would need at least 1 more condition on $a,b,c,d$ to determine them, wouldn't I? I appreciate any help. Thank you
Suppose you have one matrix $\begin{pmatrix}a&b\\c&d\end{pmatrix} = M$ such that $Mr = s$. Then it would suffice to find matrices $N_k$ in $\operatorname{SL}_2(\Bbb Z)$ such that $N_k r = Mr$. As a guess, try starting with $M$ and shifting each element by something: $$ N_k = \begin{pmatrix} a+a'&b+b'\\c+c'&d+d' \end{pmatrix}, $$ where the $a',b',c',d'$ are to be determined later. If $N_k r = Mr$, we have $$ \frac{(a + a')r + (b + b')}{(c + c')r + (d + d')} = \frac{ar + b}{cr + d}. $$ So, we want to have $a'r + b' = c'r + d' = 0$. This gives $r = -b'/a' = -d'/c'$, so let's rename a few things: write $r = -\beta/\alpha$, with $\beta,\alpha\in\Bbb Z$ (that is, we're setting $\alpha = a' = c'$, and $\beta = b' = d'$). This forces $N_k r = Mr$, so now we need to make sure our constructed $N_k$ is in $\operatorname{SL}_2(\Bbb Z)$. We need $(a + \alpha)(d + \beta) - (b + \beta)(c + \alpha) = 1$, and with a bit of work we find \begin{align*} 1&=(a + \alpha)(d + \beta) - (b + \beta)(c + \alpha)\\ &= ad + d\alpha + a\beta + \alpha\beta - (bc + c\beta + b\alpha + \alpha\beta)\\ &= ad - bc + \alpha(d - b) + \beta(a - c)\\ &= 1 + \alpha(d - b) + \beta(a - c), \end{align*} so $\alpha(d - b) = -\beta(a - c)$. By inspection, we find that $\alpha = k(a - c)$, $\beta = k(b - d)$ satisfy this for any $k\in\Bbb Z$. This gives us our infinitely many desired matrices.
Now, we need to make sure for any $r,s\in\Bbb Q$, there is some $M = \begin{pmatrix}a&b\\c&d\end{pmatrix}\in\operatorname{SL}_2(\Bbb Z)$ such that $Mr = s$. Write $r = n/m$, $s = p/q$, (both in lowest terms) with $n,m,p,q\in\Bbb Z$. \begin{align*} \begin{pmatrix}a&b\\c&d\end{pmatrix}r &=\frac{ar + b}{cr + d}\\ &= \frac{a(n/m) + b}{c(n/m) + d}\\ &=\frac{an + bm}{cn + dm}. \end{align*} Because $\gcd(n,m) = 1$, it follows that we can find $a,b$ and $c,d$ such that $an + bm = p$, $cn + dm = q$, but we need to make sure we can solve these two equations while keeping $ad - bc = 1$. Similarly to before, it is true that for any $t\in\Bbb Z$, $$ (a + tm)n + (b - tn)m = p, $$ provided $an + bm = p$, of course. Now we have a degree of freedom: our original $a,b,c,d$ didn't need to satisfy $ad - bc = 1$, but we can now try to choose $t,t'$ to force $$ (a + tm)(d - t'n) - (b - tn)(c + t'm) = 1. $$ Now we expand this, and see $$ ad + tmd - t'na - tt'mn - bc + tnc - t'mb - tt'mn = ad - bc + t(md + nc) - t'(na + mb). $$ So to guarantee we can find $t,t'\in\Bbb Z$ making this expression equal to $1$, it will be enough to show that $1 + bc - ad$ is a multiple of $\gcd(md + nc,na + mb)$. But, $\gcd(md + nc, na + mb) = \gcd(q,p) = 1$, so this is certainly possible. Thus, we have shown that given any $r,s\in\Bbb Q$, we can find a matrix $M\in\operatorname{SL}_2(\Bbb Z)$ such that $Mr = s$.