Let $Q \in \operatorname{SO}(2)$, and let $U \subseteq \mathbb R^2$ be an open, bounded, connected subset. Suppose that $QU = U$.
Is it true that $Q$ must be a disk, or the interior of a regular polygon (if $Q$ is a rotation by $2\pi/n$ then a regular $n$-gon would be invariant).
If not, can we characterize all such possible invariant $U$'s?
Edit:
As mentioned by Brian M. Scott, open annuli and their regular polygonal counterparts are also possible.
Since $Q$ is a rotation, it either has finite order $n$ or is an irrational rotation. If $Q$ is an irrational rotation and $Q(U)=U$, then $U$ is invariant under all rotations, and every connected component of $U$ is either a disk, an annulus or the whole plane.
If $Q$ has finite order, I don't think there can be a reasonnable classification. Indeed, in that case $X:=\mathbb C^*/\langle Q \rangle$ is homeomorphic to $\mathbb C^*$, and you can take any open set in $X$ and lift it to $\mathbb C$ to get a $Q$-invariant open set.
Edit: more details on the irrational case.
Observe that if $z_0 \in U$ and $U$ is $Q$-invariant (where $Q$ is an irrational rotation), then the circle $|z|=|z_0|$ is in $U$. Let $A:=\{ r \geq 0: S_r \subset U\}$, where $S_r$ is the circle of center $0$ and radius $U$. Using the previous observation, you can check that $A$ is an open subset of $\mathbb R^+$, so it's a union of intervals (open except possibly at $r=0$). Again with the observation, you can see that $U=\{ r e^{it}, (r,t) \in A \times [0,2\pi)\}$, hence a union of (at most one) disk and annuli, or the whole plane if $A=\mathbb R^+$.
