Proposition to prove. Let $\tau\colon \mathbb S^n\to \mathbb S^n$ be a conformal map, meaning that $\tau^\star g= \Lambda^2 g$ for a scalar field $\Lambda$. (Here $g$ denotes the standard metric tensor on $\mathbb S^n$). Then there exists $\theta\in\mathbb R^{n+1}, |\theta|<1$ such that $$\Lambda(\omega)=(1-\theta\cdot \omega)^{-1},\qquad \forall \omega\in \mathbb S^n.$$
I have found this statement in some sources; for example, in this paper of Chen-Frank-Weth, pag.7, "since the Jacobian determinant...", without proof.
Can you show me a proof, or point me in the right direction in the literature?
I can prove something close to the statement in the question for 2d case. Let me know if there is anything wrong!
We want to consider automorphisms of $CP^1\cong \mathbb{S}^2$. The canonical coordinate of $CP^1$ is related to the spherical coordinate through the stereographic projection: $$ z = \cot\left(\frac{\theta}{2}\right)e^{i\phi}, $$ It follows that (Stereographic projection is conformal --- from the line element) $$ g = d\theta^2 + \sin^2\theta d\phi^2 $$ $$ = \frac{4dzd\overline{z}}{(1+|z|^2)^2}, $$
The automorphisms of $CP^1$ are Möbius transformations, which can be generated by
$$ z \rightarrow az + b, \quad z \rightarrow \frac{1}{z}, $$ where $a,b\in \mathbb{C}$. But $$ \frac{4d\frac{1}{z}d\frac{1}{\overline{z}}}{(1 + |\frac{1}{z}|^2)^2} = \frac{4dzd\overline{z}}{(1+|z|^2)^2} $$ so the metric is invariant under the inverse operation, we only need to consider translation and scaling. Continuing with the transformation $z \rightarrow az + b$, $$ \frac{4|a|^2dzd\overline{z}}{(1 + |az+b|^2)^2} = \frac{|a|^2(1+|z|^2)^2}{(1 + |az + b|^2)^2} \cdot g, $$ So the conformal factor $$ \Lambda = \frac{|a|^2(1+|z|^2)^2}{(1 + |az + b|^2)^2}\\ = \frac{|a|^2}{\sin^2\left(\frac{\theta}{2}\right)(1 + |a\cot\left(\frac{\theta}{2}\right)e^{i\phi} + b|^2)^2}, $$
Denote $\omega = \begin{pmatrix}\sin\theta\cos\phi\\\sin\theta\sin\phi\\\cos\theta\end{pmatrix}$ the Cartesian coordinate of a point on the sphere, we can rewrite the expression above as $$ \Lambda(\omega) = \frac{1}{(A + \theta^T\omega)^2}, $$ where $$ A = \frac{1}{2}(1 + \frac{1+|b|^2}{|a|^2}), $$ $$ \theta_1 = \frac{1}{2}\left(\frac{b}{a} + \frac{\overline{b}}{\overline{a}}\right), \quad \theta_2 = \frac{i}{2}\left(\frac{\overline{b}}{\overline{a}}-\frac{b}{a} \right), \quad \theta_3 = \frac{1}{2}\left(1 - \frac{1 + |b|^2}{|a|^2}\right). $$