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Represent the Riemann sphere by the complex projective line, with finite $z$ represented by $[z:1]$ and infinity by $[1:0]$. An invertible matrix $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ acts by $[u:v]\mapsto [au+bv:cu+dv]$, giving the stated fractional map on finite points. Applying $B$ and then $A$ is exactly matrix multiplication $AB$, including at poles and infinity. Thereforeso $\theta$ is a group homomorphism from the special linear group to the Möbius group.

Every Möbius transformation has an invertible complex matrix representative $A$. Choose $\rho \in \mathbb{C}$ with ${\rho }^2\det A=1$. Such a nonzero square root exists in $\mathbb{C}$, and $\rho A$ has determinant one while representing the same projective map. This proves surjectivity. If the map is the identity, fixing zero and infinity forces $b=c=0$, and fixing one forces $a=d$. The determinant-one condition gives $a^2=1$. Conversely both scalar matrices $I$ and $-I$ act trivially. HenceFor the normal forms, use the fact that a complex $2\times 2$ matrix has a Jordan normal form: either it is diagonalizable or it has one nontrivial Jordan block. For a determinant-one representative, distinct eigenvalues are $\lambda ,{\lambda }^{-1}$. Diagonalizing gives $\mathrm{diag}(\lambda ,{\lambda }^{-1})$, whose projective action is $z\mapsto {\lambda }^{2}z$. Distinctness means ${\lambda }^2\ne 1$, so $\mu ={\lambda }^2$ is neither zero nor one.

If the eigenvalue is repeated, the determinant condition gives $\lambda =\pm 1$. A diagonalizable representative would be $\lambda I$ and would give the identity, excluded here. A nontrivial Jordan block is similar to $\left(\begin{array}{cc}\lambda & 1\\ 0& \lambda \end{array}\right)$, giving $z\mapsto z+1/\lambda =z\pm 1$. Any invertible conjugating matrix gives a Möbius transformation; rescaling it to determinant one does not change that map. Thus these matrix similarities are genuine conjugacies in the Möbius group, proving the requested Möbius conjugacy normal forms.

Conjugacy preserves fixed points bijectively. A nonidentity scaling has exactly the two fixed points zero and infinity, while either translation has exactly one, infinity. The identity fixes every point. This proves all three possibilities and excludes any further finite number of fixed points.

Finally, suppose $P\circ T\circ P^{-1}=S$, with $S(z)=z+\epsilon$ and $\epsilon =\pm 1$. Then $S^n(z)=z+n\epsilon \to \infty$ for every finite $z$ in the topology of the Riemann sphere; infinity remains infinity. Since $P^{-1}$ is continuous on that sphere,This is iteration of a one-fixed-point Möbius transformation. The limiting statement includes the unique fixed point itself and uses spherical convergence, which also handles intermediate iterates equal to infinity.

Work on the Riemann sphere, including zero and infinity. For $j(z)=1/z$, two iterations give the identity and one does not, so $\mathrm{ord}(j)=2$. For $k(z)=1/(1-z)$, direct composition givesNeither $k$ nor $k^2$ is the identity; in particular $0\mapsto 1\mapsto \infty \mapsto 0$. Thus

To prove the Möbius conjugacy normal forms, write a Möbius transformation as $f(z)=(az+b)/(cz+d)$ with $ad-bc\ne 0$. A finite fixed point satisfies $c{z}^2+(d-a)z-b=0$. If $c\ne 0$, this quadratic has two roots counting multiplicity, and infinity is not fixed. If $c=0$, the transformation is affine, infinity is fixed, and there is one further finite fixed point unless its slope is one. A nonidentity slope-one affine map is a nonzero translation with only infinity fixed. Thus any nonidentity Möbius transformation has one or two distinct fixed points.

If there are two, say $p,q$, choose a Möbius transformation $h$ taking them to zero and infinity. For finite $p,q$ one can use $h(z)=(z-p)/(z-q)$; the cases involving infinity use an affine map or reciprocal. The conjugate $hf{h}^{-1}$ fixes zero and infinity. Fixing infinity sets its denominator's linear coefficient to zero, and fixing zero sets its numerator's constant term to zero; hence it is $z\mapsto \mu z$ with $\mu \ne 0$.

If there is only one fixed point, send it to infinity. The conjugate has the affine form $az+b$. If $a\ne 1$, $b/(1-a)$ would be another fixed point, which is impossible; therefore $a=1,b\ne 0$. Conjugating this translation by $z\mapsto z/b$ yields $z\mapsto z+1$. The identity is already the scaling $\mu =1$. This proves the classification, and invertibility excludes $\mu =0$.

Conjugation carries fixed-point sets bijectively. The translation $z+1$ has exactly one fixed point, infinity. A nonidentity scaling has the two fixed points zero and infinity, while the identity fixes every point. Therefore $z+1$ is not conjugate to any scaling.

Finally let $f$ have finite positive order $n$. A nonzero translation has infinite order, since its $k$th iterate is $z+k$ after normalization. Thus $f$ is conjugate to $z\mapsto \mu z$, where $\mu$ has multiplicative order exactly $n$. For $n>1$, zero and infinity each form a singleton orbit. At a finite nonzero $z$, an iterate returns precisely when ${\mu }^{k}z=z$, equivalently ${\mu }^k=1$; the least positive return time is $n$. Conjugacy preserves orbit sizes. Hence the finite-order Möbius orbit sizes areFor $n>1$ exactly two points have size-one orbits; every other point has orbit size $n$.