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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 $\mathbb{R}$.

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  • $\begingroup$ "interchanging": the standard word is, I guess, "intertwining" $\endgroup$
    – YCor
    Mar 21 '19 at 20:24
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Yes.

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 $A_5$.

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    $\begingroup$ PS the bijection can be prescribed on a fundamental domain. Hence, it can be chosen piecewise isometric (with polygonal pieces, allowing segments/points). $\endgroup$
    – YCor
    Mar 21 '19 at 18:06

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