We want to find a Möbius transformation $f(z) = \frac{az+b}{cz+d}, \; a,b,c,d \in \mathbb C$ such that
$\begin{cases}z_1 \mapsto w_1 \\ z_2 \mapsto w_2 \\ \vdots \\ z_n \mapsto w_n\end{cases}$
Assuming $z \neq \frac{-d}{c}$ we get to solve the matrix equation
$\begin{pmatrix}z_1 & 1 & -w_1 z_1 & - w_1 \\ z_2 & 1 & -w_2 z_2 & - w_2 \\ \vdots & \vdots & \vdots & \vdots \\ z_n & 1 & -w_n z_n & - w_n \end{pmatrix} \mathbf x = \mathbf 0$ where $\mathbf x = \begin{pmatrix}a \\ b \\ c\\ d \end{pmatrix}$
For $f$ to be a function, we need to assume that $\left(z_i \mapsto w_i \right) \land \left(z_i \mapsto w_j \right)$ aren't true whenever $w_i \neq w_j$ (that is, every element in the domain is mapped to exactly one element in the range). Under this assumption, what can we say about the solutions to the matrix equation? When do they exist and what can we say about them?
Note that the trivial solution $\mathbf x = \mathbf 0$ means a Möbius transformation with our constraints cannot exist because $\mathbf x = \mathbf 0 \implies$ "$f(z) = \frac{0}0$".