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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.