**Step 1: Conjugation preserves fixed point count.** [claim:Conjugation Preserves Fixed Points] If $f = g^{-1}hg$ for $f, g, h \in \mathcal{M}$, then $\alpha$ is a fixed point of $f$ if and only if $g(\alpha)$ is a fixed point of $h$. [/claim] [proof] $f(\alpha) = \alpha \iff g^{-1}hg(\alpha) = \alpha \iff hg(\alpha) = g(\alpha)$, i.e., $g(\alpha)$ is fixed by $h$. Since $g$ is a bijection of $\mathbb{C}_\infty$, the correspondence $\alpha \leftrightarrow g(\alpha)$ is a bijection between the fixed points of $f$ and the fixed points of $h$. [/proof] **Step 2: Count fixed points of the standard forms.** By the [Conjugacy Classes in the Möbius Group](/theorems/812) theorem, every non-identity $f$ is conjugate to one of: - $z \mapsto az$ with $a \neq 0, 1$: the fixed points are $z = 0$ and $z = \infty$ (solving $az = z$ gives $z(a - 1) = 0$). This gives exactly $2$ fixed points. - $z \mapsto z + 1$: the only fixed point is $z = \infty$ (solving $z + 1 = z$ has no finite solution). This gives exactly $1$ fixed point. **Step 3: Conclude.** By Step 1, $f$ has the same number of fixed points as its conjugate standard form. So every non-identity Möbius map has either $1$ or $2$ fixed points.