The inversion $I:M\rar M$, $x\mapsto x/\Vert x\Vert^2$ is conformal on $M\colon=\R^n\sm\{0\}$ with scaling function $h(x)=1/\Vert x\Vert^2$. Hint: $DI(x)v=(v-2x\la x,v\ra/\Vert x\Vert^2)/\Vert x\Vert^2$.
Since the derivative of $\Vert x\Vert^2$ in the direction $v$ is $2\la x,v\ra$ we get by the product rule:
$$
DI(x)v
=v\Vert x\Vert^{-2}-2x\la x,v\ra\Vert x\Vert^{-4}
=(v-2x\la x,v\ra/\Vert x\Vert^2)/\Vert x\Vert^2~.
$$
The map $R:v\mapsto v-2\la x,v\ra x/\Vert x\Vert^2$ is the reflection about the hyperplane orthogonal to $x$, thus $R$ is an isometry and since $DI(x)=\Vert x\Vert^{-2}R$ it follows that:
$$
DI(x)^*DI(x)=R^*R/\Vert x\Vert^{-4}=\Vert x\Vert^{-4}~.
$$
Hence $I$ is conformal with scaling function $h(x)=\Vert x\Vert^{-2}$.