In your question you've taken the condition "$M(C)=C$ " and misunderstood it to mean that it leaves every single point on the circle unchanged, so that $M(z_{1,2,3})=z_{1,2,3}$ respectively. This interpretation is false. I provided a counterexample in my comment, but you misunderstood the point of the comment: the map $z\mapsto iz$ is a Möbius transformation that maps the unit circle to the unit circle (hence it satisfies the condition $M(C)=C$) however $M(z)=z$ is false for every single particular element $z$ on the unit circle. The point is that your interpretation of the condition does not make sense so you cannot build an approach off of it. I am not saying that pure rotations are the only transformations preserving circles; they are not.
The key to understanding this problem is geometry - more specifically, spherical geometry. We will first prove a particular case of the proposition, and then show this entails the full proposition.
Lemma. For any $u,v\in\mathbb{C}$, there is a Möbius transformation $\varphi$ such that $\varphi(u)=v$ and $\varphi(\mathbb{R})=\mathbb{R}$.
Visual Proof. The isometries of the real line are affine transformations $x\mapsto ax+b$, so we need to see that such a transformation is sufficient. Translations have an obvious interpretation of moving everything horizontally but not vertically, while real dilations will act on complex $z$ as dilations of $z$ on the ray extending from the origin to $z$. The line between the origin and $u$ will take on every possible $y$ (imaginary) value, so we can perform a real dilation on $u$ until it matches $v$ in height; on top of that we just need to translate the appropriate horizontal distance to get to $v$. With these two actions composed together we have our affine transformation, as desired.
Proposition. For any (generalized) circle $C$ and $u,v\in\mathbb{C}-C$, there is a Möbius transformation $\varphi$ such that $\varphi(u)=v$ and $\varphi(C)=C$.
Proof. Fix a Möbius transformation $\psi$ that maps $C\to\mathbb{R}$. By our lemma there is a transformation that preserves the real line but maps $\psi(u)$ to $\psi(v)$; call it $\rho$. Then our desired map is $\psi^{-1}\circ\rho\circ\psi $.
This write-up requires a bit of expansion in order to be as comprehensive as I'd like it to be, and I'll get to it next thing I do on MSE, but right now I have to go to sleep and catch some ZZZ's! (Also, this is basically of an elaboration on Blatter's answer and clarification of my comment.)