= Solution
Represent the <Riemann sphere> by the complex <projective line>, with finite $z$ represented by $[z:1]$ and infinity by $[1:0]$. An invertible <matrix> $A=\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)$ 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. Therefore
$$
\theta(AB)=\theta(A)\circ\theta(B),
$$
so $\theta$ 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. Hence
$$
\boxed{\ker\theta=\{I,-I\},\qquad
\mathcal M\cong SL_2(\mathbb C)/\{I,-I\}.}
$$
For the normal forms, use the fact that a complex $2\times2$ <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 $\operatorname{diag}(\lambda,\lambda^{-1})$, whose projective action is $z\mapsto\lambda^2z$. Distinctness means $\lambda^2\ne1$, so $\mu=\lambda^2$ is neither zero nor one.
If the <eigenvalue> is repeated, the <determinant> condition gives $\lambda=\pm1$. A diagonalizable representative would be $\lambda I$ and would give the identity, excluded here. A nontrivial <Jordan block> is similar to $\left(\begin{smallmatrix}\lambda&1\\0&\lambda\end{smallmatrix}\right)$, giving $z\mapsto z+1/\lambda=z\pm1$. 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 $S(z)=z+\varepsilon$ and $\varepsilon=\pm1$. Then $S^n(z)=z+n\varepsilon\to\infty$ for every finite $z$ in the topology of the <Riemann sphere>; infinity remains infinity. Since $P^{-1}$ is continuous on that <sphere>,
$$
\boxed{T^n(z)=P^{-1}\bigl(S^n(P(z))\bigr)\longrightarrow
P^{-1}(\infty)=z_0\quad\text{for every }z.}
$$
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.
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