Can a weaker version of the Hausdorff paradox be proved without AC?

Question

The Hausdorff paradox is an incredibly counter-intuitive consequence of the axiom of choice; it is also important for demonstrating the non-existence, under AC, of a rotation-invariant measure on the discrete $\sigma$-algebra of the sphere.

But there is a part of the Hausdorff paradox that seems consistent with the existence of a rotation-invariant measure on the discrete $\sigma$-algebra, and yet still seems (to me) very counter-intuitive; so I am curious as to whether it can be proved without AC:

Is it a theorem of "ZF + Dependent Choice" that there exists a non-empty set $A \subset \mathbb{S}^2$ and rotations $R,R' \colon \mathbb{S}^2 \to \mathbb{S}^2$ such that $R(A)$ is disjoint from $A$ and $R'(A)=A \cup R(A)\,$?

Is it, at least, a theorem of "ZF + Dependent Choice" that there exists a non-empty set $A \subset \mathbb{S}^2$ and a rotation $R \colon \mathbb{S}^2 \to \mathbb{S}^2$ such that $A \subsetneq R(A)\,$?

(I fear there's some really obvious construction of such a set $A$ that I've missed - but I'll take the risk and ask the question!)

Answer 1

Following the suggestion in the first comment below my question (and with the help of the second comment), I can give an example of a scenario that is "even worse" than what I requested, where $A \cup R(A)$ is a proper subset of $R'(A)$.

(Finding an example where we actually have equality is proving remarkably difficult to me, so I still invite answers regarding having equality, i.e. $A \cup R(A)=R'(A)$.)

Regard $\mathbb{S}^2$ as the unit sphere about the origin in $\mathbb{R}^3$.

Fix $\alpha \in \mathbb{R}$ such that $\cos(\alpha)$ is transcendental, and define the matrices \begin{align*} M_1 \ &= \ \left( \begin{array}{c c c} \cos(\alpha) & -\sin(\alpha) & 0 \\ \sin(\alpha) & \cos(\alpha) & 0 \\ 0 & 0 & 1 \end{array} \right) \\ & \\ M_2 \ &= \ \left( \begin{array}{c c c} \cos(\alpha) & 0 & -\sin(\alpha) \\ 0 & 1 & 0 \\ \sin(\alpha) & 0 & \cos(\alpha) \end{array} \right). \end{align*}

According to p227 of here, since $\cos(\alpha)$ is transcendental, $\{M_1,M_2\}$ forms a generating set of a free group contained in $\mathrm{SO}(3)$; and by Euler's rotation theorem, each non-trivial member of the group $G$ generated by $\{M_1,M_2\}$ fixes exactly 2 points. Hence (since $\mathbb{S}^2$ is uncountable) there exists a point $x^\ast \in \mathbb{S}^2$ which is not fixed under any non-trivial member of $G$.

Let $G^+$ be the monoid generated by $\{M_1,M_2\}$, and take \begin{align*} A \ &= \ \{M_1Mx^\ast : M \in G^+\} \\ R \ &= \ M_2M_1^{-1} \\ R' \ &= \ M_1^{-1}. \end{align*} Then $R(A)=\{M_2Mx^\ast : M \in G^+\}$ and $R'(A)=\{Mx^\ast : M \in G^+\}$. So $A$ is disjoint from $R(A)$, and $$ R'(A) \ = \ A \cup R(A) \cup \{x^\ast\}. $$