= Solution
Represent a <Möbius transformation> by a matrix, with nonzero scalar multiples representing the same map. Matrix multiplication reproduces composition, so $\theta$ is a <group homomorphism>. Every Möbius map has a representing matrix $A\in GL_2(\mathbb C)$; multiplying $A$ by either square root of $(\det A)^{-1}$ gives a matrix in $SL_2(\mathbb C)$ representing the same map. Thus $\theta$ is surjective. A matrix in its kernel represents the identity and is therefore scalar; determinant one leaves
$$
\boxed{\ker\theta=\{I,-I\}}.
$$
If a nonidentity Möbius map $T$ has two distinct fixed points, conjugate those points to $0$ and $\infty$. The conjugated map fixes both and hence has the form $S(z)=\mu z$ with $\mu\ne0,1$. If $T$ has one fixed point, conjugate it to $\infty$. The resulting affine map has the form $az+b$; having no finite fixed point forces $a=1$ and $b\ne0$, and conjugating by a scaling makes it $S(z)=z+1$.
Directly, the finite fixed points satisfy
$$
cz^2+(d-a)z-b=0.
$$
Including $\infty$ when appropriate, a nonidentity Möbius map consequently has one or two distinct fixed points.
Finally suppose $T$ has the unique fixed point $z_0$. With the conjugacy $R$ above,
$$
RTR^{-1}(z)=z+1,
\qquad
(RTR^{-1})^n(z)=z+n.
$$
Every finite point tends to $\infty$ on the <Riemann sphere>, while $\infty$ itself is fixed. Applying the continuous map $R^{-1}$ shows that
$$
\boxed{T^n(z)\longrightarrow z_0}
$$
for every $z\in\mathbb C\cup\{\infty\}$.
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