Assume g,f:A⊂RM→S2 are two bijective functions defined on the set A. Now assume a constraint C: ∀x,y∈A,∃R∈SO(3):Rf(x)=f(y)⟺Rg(x)=g(y). Does C imply f=±g?
[Update]: The answer is "yes" when C is modified to ∀x,y∈A,∀R∈SO(3):Rf(x)=f(y)⟺Rg(x)=g(y).
Assume g,f:A⊂RM→S2 are two bijective functions defined on the set A. Now assume a constraint C: ∀x,y∈A,∃R∈SO(3):Rf(x)=f(y)⟺Rg(x)=g(y). Does C imply f=±g?
[Update]: The answer is "yes" when C is modified to ∀x,y∈A,∀R∈SO(3):Rf(x)=f(y)⟺Rg(x)=g(y).
You are correct, if you change C so R is \forall-quantified, then C implies f = \pm g.
Pairs \{v, -v\} have distinct stabilizers under SO(3) so setting x = y we see f = \pm g pointwise. But then if f(x) = g(x) and f(y) = -g(y), picking any R such that Rf(x) = f(y), if we were to have Rg(x) = g(y) then we would have Rf(x) = Rg(x) = g(y) = -f(y) = -Rf(x), a contradiction. Of course neither the choice of domain nor the fact the dimension of the sphere is 2 matters.
(Your argument is probably correct as well, but I had already written this when I realized \cdot means dot product.)