Is there a natural or canonical map

Question

Is there any natural or canonical map $$ \begin{array}{c} f:S^{2}\times S^{2}\times S^{2}\rightarrow S^{2}\times S^{2}\times S^{2}\\ f:\left(p_{1},p_{2},p_{3}\right)\mapsto\left(p'_{1},p'_{2},p'_{3}\right), \end{array} $$ where $S^{2}\subset\mathbb{R}^{3}$ is the unit sphere of $\mathbb{R}^{3}$ with the Euclidean inner product, that is some type of anti-conformal in the sense that, for example, when $p_{1}=p_{2}=p_{3}$ one has that $p'_{1},p'_{2},p'_{3}$ are orthogonal, and when $p_{1},p{}_{2},p{}_{3}$ are orthogonal one has that $p_{1}=p_{2}=p_{3}$? Thanks a lot.

Answer 1

I think such a map cannot exist for topological reasons, at least if you want your map $f$ to be a homeomorphism (and if I have understood the question correctly).

The subset $A = \{(x,y,z) \in S^2 \times S^2 \times S^2 | x=y=z\}$ is homeomorphic to $S^{2}$. The subset $B = \{(x,y,x)| $x,y,z$ \text{ are pairwise orthogonal}\}$ is three dimensional- for each $x$ there is a circle of points orthogonal to $x$ in which $y$ and $z$ can vary (whilst still being orthogonal). This shows $B$ is homeomorphic to an $S^{1}$-bundle over $S^{2}$. Hence, since $A$ and $B$ are not homeomorphic, there cannot be a homeomorphism mapping $A$ to $B$.