Assume $g,f\colon A\subset\mathbb{R}^M\rightarrow\mathbb{S}^2$ are two bijective functions defined on the set $A$. Now assume a constraint $C$: $\forall x,y\in A, \exists R\in SO(3)\colon Rf(x)=f(y)\iff Rg(x)=g(y)$. Does $C$ imply $f=\pm g$?

[Update]: The answer is "yes" when $C$ is modified to $\forall x,y\in A, \forall R\in SO(3)\colon Rf(x)=f(y)\iff Rg(x)=g(y)$.