I apologize in advance if this question is so trivial or too low level.
Let $\Gamma$ be a Fuchsian group. Let $\mathcal{F}$ be the set of pairs $(\mu,f)$, where $\mu \in L^\infty(\mathbb{C})$ such that $\mu(\overline{z})=\overline{\mu(z)}$, $||\mu|| < 1$, and $f$ is a quasiconformal mapping of the plane satisfying beltrami differential equation with beltrami coefficient $\mu$:
\begin{equation} \mu f_z = f_{\overline{z}} \end{equation}
The solution exists up to a Mobius transformation.
Let $\mathcal{F}(\Gamma)$ pairs $(\mu,f) \in \mathcal{F}$ be such that $\mu \circ \gamma \frac{\overline{\gamma'}}{\gamma'} = \mu$ for all $\gamma \in \Gamma$. In this case, $\Gamma_f = f\circ \Gamma \circ f^{-1}$ is a Fuchsian group.
I'm wondering when $\Gamma = f\circ \Gamma \circ f^{-1}$ would hold. (Here, by "=" I mean equal as subsets of $PSL(2,\mathbb{R})$.
What I do know are some special (trivial) cases.
1) $\Gamma = 1$. No conditions on $f$ needs to be imposed.
2) $\Gamma$ is generated by a single parabolic element $g$. Then, the following conditions are sufficient: a) $f$ fixes the fixed point of $g$. b) $f$ fixes $b$ and $g(b)$ for some other point $b$.
3) $\Gamma$ is generated by a single hyperbolic element $g$. Then, the following conditions are sufficient: a) $f$ fixes the fixed points of $g$. b) $f$ has some other fixed point.
But outside of these cases, I don't have the slightest clue.
