Bijection from $S^2$ to itself interchanging actions of $A_5$

Let $$X$$ and $$Y$$ be two copies of $$S^2$$, and let $$A_5$$ act on each of them (as a group of rotations). Call these actions $$\theta_X$$ and $$\theta_Y$$.

Moreover, let $$g \in A_5$$ be a fixed element of order $$5$$ (without loss of generality, the cycle $$(12345)$$) and suppose that $$\theta_X(g)$$ is a rotation by $$\frac{2 \pi}{5}$$ and $$\theta_Y(g)$$ is a rotation by $$\frac{4 \pi}{5}$$. These actions are unique (up to rotations of $$X$$ and $$Y$$) and correspond to the two distinct irreducible representations of $$A_5$$ with dimension $$3$$. Consequently, the following question is well defined:

Does there exist a bijection $$f : X \rightarrow Y$$ which 'commutes with' these actions, in the sense that:

$$\theta_Y(h) \circ f = f \circ \theta_X(h)$$

for all elements $$h \in A_5$$?

Note that if we replace $$S^2$$ with the (dense) intersection $$S^2 \cap \mathbb{Q}[\sqrt{5}]^3$$, then there is such a bijection: the field automorphism which interchanges $$\sqrt{5}$$ and $$-\sqrt{5}$$. Unfortunately, this does not extend to a field automorphism of $$\mathbb{R}$$, as there are no nontrivial field automorphisms of $$\mathbb{R}$$.

• "interchanging": the standard word is, I guess, "intertwining"
– YCor
Mar 21, 2019 at 20:24

You need to describe $$X$$ and $$Y$$ as $$A_5$$-sets. One is obtained from the other by twisting the action by a non-inner automorphism $$\sigma$$ of $$A_5$$, say $$Y=X^\sigma$$. But one observation is that two subgroups of $$A_5$$ are conjugate in $$A_5$$ iff they're conjugate in $$S_5$$. So any $$A_5$$-set $$X$$ is isomorphic to $$X^\sigma$$. This applies to this question.
Addendum: although it was unnecessary to describe the sphere as $$A_5$$-set, this is easy. The stabilizer of a generic point (= not vertex, center of face or edge) is trivial. The 12 vertices form an orbit, with stabilizer a 5-Sylow. The 20 face centers form an orbit, with stabilizer a 3-Sylow. The 30 edge centers form an orbit, with stabilizer a cyclic subgroup of order 2. Thus, as $$A_5$$-set, the sphere is $$\mathfrak{c}\cdot A_5 \,\sqcup \,A_5/2\,\sqcup \,A_5/3\,\sqcup \,A_5/5,$$ where $$A_5/i$$ denotes the quotient by a subgroup of order $$i$$, which for each of $$i=2,3,5$$, is unique up to conjugacy in $$A_5$$.