Relation between two homomorphisms from $SO(3)$ to the Möbius group $PGL(2,\mathbb{C})$ Let $f, g$ be two homomorphisms from $\mathrm{SO}(3)$ to $PGL(2,\mathbb{C})$. Does there exist $S \in \mathrm{PGL}(2,\mathbb{C})$ such that for all $X\in \mathrm{SO}(3)$ we have $g(X)=Sf(X)S^{-1}$?
Background: I have images which are stereographic projections of subparts of a sphere. According to Wikipedia the images transform into each other using Möbius transforms. I want to parametrize these transforms in terms of the rotations given by the actuator. If the above is true the parametrization would be quite simple: i can pick any homomorphism and then just need to parametrize $S$.
 A: Any continuous homomorphism $SO_3 \to SL_2(\mathbb C)$ is trivial because $SO_3$ is compact and has only one irreducible representation of dimension at most two: the trivial representation.
EDIT: the revised question may still be answered by representation theory. A continuous homomorphism $f: SO_3 \to PGL_2(\mathbb C)$ has a unique lift to a homomorphism $\tilde f: SU_2 \to SL_2(\mathbb C)$. By the representation theory of $SU_2$, there are two representations of dimension two: the trivial representation and the defining representation on $\mathbb C^2$. Thus if $f,g: SO_3 \to PGL_2(\mathbb C)$ are both nontrivial and continuous, their lifts $\tilde f, \tilde g$ are conjugate and thus $f$ and $g$ are conjugate.
A: There are two continuous group morphisms $SO_3\to PSL_2\mathbb{C}$ up to conjugacy: the obvious one as rotations of the Riemann sphere, and the trivial one with image the identity element. The proof: We see from the Lie algebra of $SO_3$ (cross product of vectors in $\mathbb{R}^3$) that $SO_3$ has simple Lie algebra. Hence the image $G$ of any morphism is a 0-dimensional or a 3-dimensional compact and connected group. If 0-dimensional, it is the identity element, because it is connected. So suppose that $G$ is 3-dimensional.
By $QR$ decomposition, $PSL_2\mathbb{C}$ retracts to $SO_3$, so $SO_3$ is a maximal compact subgroup, and is connected. All maximal compact subgroups are conjugate. So any maximal compact subgroup containing $G$ is a conjugate of $SO_3$. By dimension count, and connectivity, $G$ is a maximal compact subgroup, so a conjugate of $SO_3$, so unique up to conjugacy. We thus reduce to the problem of proving that $SO_3$ automorphisms are inner, which is clear from its Dynkin diagram, as it is rank one, so has a Dynkin diagram with only one node, hence no noninner automorphisms.
