Can a weaker version of the Hausdorff paradox be proved without AC? 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!)
 A: 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
$$ R'(A) \ = \ A \cup R(A) \cup \{x^\ast\}. $$
