Let $A=\{0,1,\infty,a_1,\ldots,a_n\}$ and $B=\{0,1,\infty,b_1,\ldots,b_n\}$ be subsets of the Riemann sphere.
Let $\sigma$ be an automorphism of the Riemann sphere, i.e., a Möbius transformation, such that $\sigma(A) = B$.
What can one say about $\sigma$?
Example. Suppose that $n=0$. Then $\sigma$ is an automorphism sending $\{0,1,\infty\}$ to $\{0,1,\infty\}$. So $\sigma$ is either the identity map or $z\mapsto \frac{1}{z}$.
Example. Suppose that $n=1$. Under the hypothesis, we have that $b_1 = a_1$. (The cross ratio of $A$ and $B$ should be equal.) So there should be four possibilities for $\sigma$; one of them being the identity map.
Three points, $(a,\sigma(a)), (b,\sigma(b)), (c,\sigma(c))\in\mathbb{C}^2$, are enough to wholly determine a Möbius transformation; adding in any more information than this will create an overdetermined situation.
Also, if you have $\{a,b,c\}$ and $\{\sigma(a),\sigma(b),\sigma(c)\}$ without knowledge of which argument goes to what output, you have a total of $3!=6$ different transformations. By making the coefficients of the Möbius transformation unknowns and multiplying through by the denominator, we can solve for the transformation exactly via linear algebra (given a known which-goes-to-what correspondence). Wikipedia has how with an explicit determinant formula. You can then say anything you want about $\sigma$ once you know what precisely it can be.