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 $\mathbb{R}$.



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




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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 $A\_5$.


