Suppose $|v| < 1$ and $\Vert z\Vert=1$. Is the classical aberration map $F_c:S^n\rar S^n$,
$$
x\mapsto\frac{x-vz}{\Vert x-vz\Vert}
$$
a conformal transformation of $S^n$? 2. Compute the inverse of $F_c$.
$F_c$ is the composition of the translation $T(x)=x-vz$ and the map $G(y)=y/\Vert y\Vert$. Since
$$
DG(y)u
=\frac u{\Vert y\Vert}-\frac{y\la y,u\ra}{\Vert y\Vert^3}
=\frac1{\Vert y\Vert}\Big(u-\frac{\la y,u\ra}{\Vert y\Vert^2}\,y\Big)
$$
and $DT(x)u=u$ we get by the chain rule:
$$
DF_c(x)u
=DG(T(x))DT(x)u
=DG(x-vz)u
=\frac1{\Vert x-vz\Vert}\Big(u-\frac{\la x-vz,u\ra}{\Vert x-vz\Vert^2}\,(x-vz)\Big)
$$
In particular for $x\in S^n$ and $u\perp x$:
$$
DF_c(x)u
=\frac1{\Vert x-vz\Vert}\Big(u+\frac{\la vz,u\ra}{\Vert x-vz\Vert^2}\,(x-vz)\Big)
$$
This implies that:
\begin{eqnarray*}
\Vert x-vz\Vert^2\la DF_c(x)u,DF_c(x)u\ra
&=&\Vert u\Vert^2
+\frac{\la vz,u\ra^2}{\Vert x-vz\Vert^4}\Vert x-vz\Vert^2
+2\frac{\la vz,u\ra}{\Vert x-vz\Vert^2}\la u,x-vz\ra\\
&=&\Vert u\Vert^2
+\frac{\la vz,u\ra^2}{\Vert x-vz\Vert^2}
-2\frac{\la vz,u\ra^2}{\Vert x-vz\Vert^2}\\
&=&\Vert u\Vert^2
-\frac{\la vz,u\ra^2}{\Vert x-vz\Vert^2}~.
\end{eqnarray*}
Thus $F_c:S^n\rar S^n$ is not conformal.
2. As for the inverse we need to solve the equation
$$
\frac{x-vz}{\Vert x-vz\Vert}=y
$$
for given $y\in S^n$. Now $x=y\Vert x-vz\Vert+vz$ and since $x\in S^n$:
$$
\Vert x-vz\Vert^2+2v\la z,y\ra\Vert x-vz\Vert-1+v^2=0~.
$$
Solving for $\Vert x-vz\Vert$ we get
$$
\Vert x-vz\Vert=-v\la z,y\ra+\sqrt{v^2\la z,y\ra^2+1-v^2}
$$
It follows that
$$
x=y\Big(-v\la z,y\ra+\sqrt{v^2\la z,y\ra^2+1-v^2}\Big)+vz
$$