$\textbf{Background for question}$: Let $L_{k} = \{z \in \mathbb{C} : \arg(z) = \frac{\pi k}{n} \}$, and set $\Phi$ equal to the group of compositions of reflections in the lines $L_{k}, k=1,...,2n$.
A decomposition (i.e. $P_{k}$ are interior disjoint closed sectors, and $\bigcup_{k=1}^{N} P_{k} = \mathbb{C}$) a of the plane into closed sectors $\{P_{k} \}_{k=1}^{N}$ will be called $\textbf{symmetric}$ if for any $\phi \in \Phi$, $\{ \phi(P_{k}) \}_{k=1}^{N} = \{P_{k} \}_{k=1}^{N}$. Fix such a symmetric decomposition $\{P_{k} \}_{k=1}^{N}$.
Suppose $G$ is an open, connected subset of the plane which satisfies $\phi(G) = G$ (say $G$ is $\Phi$-invariant) , for any $\phi \in \Phi$.
Let $\{\lambda_{k} \}_{k=1}^{N}$ be a collection of rotations (about $0$) for which $\{\lambda_{k}(P_{k}) = S_{k} \}_{k=1}^{N}$ is also a decomposition of the plane into interior disjoint closed sectors, which satisfies the following property:
If $S_{k} \cap S_{l} \neq \emptyset$, then there exists some $\phi \in \Phi$ for which $\phi(\lambda_{k}^{-1}(S_{k} \cap S_{l})) = \lambda_{l}^{-1}(S_{k} \cap S_{l})$.
Define $\tilde{G} = \bigcup_{k=1}^{N} \lambda_{k}(P_{k} \cap G)$, this is necessarily an open set (one can check this).
Now suppose $u : G \rightarrow \mathbb{R}$ is harmonic and $\Phi$-invariant, i.e. $u \circ \phi = u$ for all $\phi \in \Phi$. Define $\tilde{u}$ on $\tilde{G}$, by setting it equal to $u \circ \lambda_{k}^{-1}$ on $\lambda_{k}(G \cap P_{k})$. Suppose $\tilde{u}$ is harmonic on $\tilde{G}$.
$\textbf{Question}$:
Suppose $\tilde{u}$ is harmonic on $\tilde{G}$, and denote by $\tilde{G}_{k}$ a connnected component of $\tilde{G}$ containing a connected component of the interior of $\lambda_{k}(G \cap P_{k}) \neq \emptyset$. Apparently, by the identity theorem for harmonic functions, since $\tilde{u}$ and $u \circ \lambda_{k}^{-1}$ coincide on a connected component of the interior of $\lambda_{k}(G \cap P_{k})$, they coincide on $\tilde{G}_{k}$, and then we must also have $\tilde{G}_{k} = \lambda_{k}(G)$.
Where I am lost is the last part of this argument. Why must $\tilde{G}_{k} = \lambda_{k}(G)$?