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](https://en.m.wikipedia.org/wiki/M%C3%B6bius_transformation) 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 $S$.



Relation between two homomorphisms from $SO(3)$ to the Möbius group $PGL(2,\mathbb{C})$
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.


