Let $\phi(z)=\frac{az+b}{cz+d}$ be a Möbius transformation, with $a,b,c,d\in \mathbb{R}$ and $ad-bc=1$ (so $\phi$ can be viewed as an element of $SL_2(\mathbb{R})$. Let $\mathbb{H}=\{z\in\mathbb{C}: \Im(z)>0\}$ be the upper half-plane, and fix a compact set $K\subset \mathbb{H}$. Suppose $z,\phi(z)\in K$. Can we provide a bound for $a,b,c,d$ depending only on the set $K$?
Since $K\subset\mathbb{H}$, in particular $K\cap \mathbb{R}=\emptyset$, so $K$ is at a positive distance from the real axis. Since it is also bounded, we deduce the existence of positive numbers $r$ and $M$ such that the inequalities
$$ r \le |w|\le M$$ $$r\le \Im (w)\le M$$
hold for every $w\in K$, and so they hold in particular for $z$ and $\phi(z)$. I've tried playing around with this but got nowhere. I also tried to use the fact that $$\Im \phi(z)=\frac{\Im z}{|cz+d|^2}$$ but I keep getting bounds along the lines of $(|cz|-|d|)^2\le \frac{M}{r}$ which I can't use to dominate $c$ or $d$. I've been able to solve the related problem $|\phi(i)-i|\le \varepsilon \implies |a|,|b|,|c|,|d|\le C_\varepsilon$, where the computations are much simpler, and tried using adequate Möbius maps to transfer the general case to this one, but couldn't get it to work.
EDIT: I was able to get
$$r\le |\Im \phi(z)|=\frac{|\Im z|}{|cz+d|^2}\le \frac{M}{c\Im z}\le \frac{M}{c^2r^2}$$
and so $c$ is bounded. But then
$$r\le |\Im \phi(z)|\le \frac{M}{|\Re (cz+d)|^2}=\frac{M}{|d+c\Re z|^2}$$
and thus $|d+c\Re z|\le \left(\frac{M}{r}\right)^{1/2}$, so $d$ must be bounded, since $c$ and $|z|$ are.
I still haven't been able to control $a$ and $b$. Getting lower bounds for $|c|$ and $|d|$ might be useful, since we know $ad-bc=1$, but I don't know if that is possible.
Let's first consider the case of $z=i \in K.$ Then
$$\tag 1 \phi(i) = \frac{ai+b}{ci+d} = \frac{ac+bd+i(ad-bc)}{c^2+d^2} = \frac{ac+bd+i}{c^2+d^2}.$$
So the imaginary part of $\phi(i)$ equals $1/(c^2+d^2).$ Because $K$ is compact, the imaginary parts of points in $K$ must be bounded above and below by positive constants (depending on $K$). This implies $c^2+d^2$ is also bounded above and below by positive constants depending on $K$.
On the other hand,
$$|\phi(i)| =\frac{|ai+b|}{|ci+d||}= \frac{ (a^2+b^2)^{1/2} }{ (c^2+d^2)^{1/2} }.$$
Since $|\phi(i)|$ is itself bounded above and below by positive constants (again depending on $K$), the same must be true of the numerator $(a^2+b^2)^{1/2},$ since we've already shown this for the denominator.
We're done in the case $z=i.$ But I think the general case is reducible to this case. After all, whatever $z,K$ are, we are just off by a translation and dilation, the former bounded and the latter bounded above and below by positive constants (again depending on $K$).