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 $f=\pm g$?

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 $\forall x,y\in A, \forall R\in SO(3)\colon Rf(x)=f(y)\iff Rg(x)=g(y)$.

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 $f=\pm g$?

$C$ implies $\forall x,y\in A, f(x)g(x)=f(y)g(y)$ which is not isometry but similar. If I could prove transformation between g and f is isometric i.e. $\forall x,y\in A, f(x)f(y)=g(x)g(y)$ it's not hard to show $f=\pm g$. Do I need more constraint on f and g to imply isometry?

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 $f=\pm g$?