This is what I came up with if $C$ is modified to $\forall x,y\in A, \forall R\in SO(3): Rf(x)=f(y)\iff Rg(x)=g(y)$. Let $\theta(R)$ denote the rotation angle corresponding to rotation matrix $R$. We can show that $f(x)\cdot f(y)=\max\limits_{R\in SO(3): Rf(x)=f(y)} \cos{\theta(R)}$. This implies that if $C$ holds $\forall x,y\in A f(x)f(y)=g(x)g(y)$ and consequentially we can show that $\exists R'\in RO(3): g=R'f$. By plugging this into C this implies that $\forall x,y\in A, R\in SO(3): Rf(x)=f(y) \iff RR'f(x)=R'f(y)$ which implies $\forall R\in SO(3): RR'=R'R$ equivalent to $R'=\pm I$.

My original notation might be wrong. What I meant was that if for any $R\in SO(3)$ for which $Rf(x)=f(y)$ holds $Rg(x)=g(y)$ also holds and vice versa. I think I have to change $\exists R$ to $\forall R$. Right? I removed the second paragraph for the sake of clarity.