The strategy is to use the surjective homomorphism $\Phi : \mathrm{GL}_2(\mathbb{C}) \to \mathcal{M}$ from the [Möbius Group as Matrix Quotient](/theorems/809) theorem. Conjugacy in $\mathrm{GL}_2(\mathbb{C})$ descends to conjugacy in $\mathcal{M}$, so it suffices to classify $2 \times 2$ matrices up to conjugation and then compute the corresponding Möbius maps. **Step 1: Conjugacy descends through $\Phi$.** If $P A P^{-1} = B$ in $\mathrm{GL}_2(\mathbb{C})$, then $\Phi(P)\Phi(A)\Phi(P)^{-1} = \Phi(B)$ in $\mathcal{M}$. So conjugate matrices give conjugate Möbius maps. **Step 2: [Jordan normal form](/theorems/864) cases.** Every $A \in \mathrm{GL}_2(\mathbb{C})$ is conjugate to one of: (i) $\begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix}$ with $\lambda_1 \neq \lambda_2$: this gives $\Phi(A)(z) = \frac{\lambda_1}{\lambda_2} z = az$ with $a = \lambda_1/\lambda_2 \neq 0, 1$. (ii) $\begin{pmatrix} \lambda & 0 \\ 0 & \lambda \end{pmatrix}$: this gives $\Phi(A)(z) = z$, the identity map. (iii) $\begin{pmatrix} \lambda & 1 \\ 0 & \lambda \end{pmatrix}$: this gives $\Phi(A)(z) = \frac{\lambda z + 1}{\lambda} = z + \frac{1}{\lambda}$. A further conjugation by $z \mapsto \lambda z$ shows this is conjugate to $z \mapsto z + 1$. **Step 3: Conclude.** Every $f \in \mathcal{M}$ is conjugate to $z \mapsto z$ (identity), $z \mapsto z + 1$, or $z \mapsto az$ for some $a \in \mathbb{C} \setminus \{0, 1\}$.