Do identical orbit tiles imply identical combinatorial types?

Given a periodic trajectory on a triangle, we can associate to this trajectory a sequence of integers $1,2$ and $3$ by labeling the edges of the triangle and taking the sequence of edges the trajectory strikes. It is natural to consider the resulting free necklace, as this makes the result independent of our choice of parametrization of the trajectory (including reversing it). We call this the combinatorial type of the trajectory. Given a combinatorial type $C$, we can associate to $C$ the set of triangles which admit periodic trajectories with combinatorial type $C$, which we call the orbit tile of $C$ and denote $\mathcal O(C)$. Regarding the space of triangles as the square $(0,\pi/2)\times (0,\pi/2)$ (a point is understood as specifying two of the angles of the triangle, determining it up to similarity), orbit tiles are always the intersection of an open set bounded by analytic curves with some affine subspace of the plane.

Suppose $\mathcal O(C)=\mathcal O(C')\neq \emptyset$. Does it follow that $C=C'$ (as free necklaces)?

Clearly the hypothesis that the orbit tiles be nonempty is necessary. I'm reasonably sure a careful analysis of the curves bounding the orbit tile will prove this when the orbit tile is an open set. However, I'm not sure of the case where $\mathcal O(C)$ is the intersection of an open set and a line, and I would not be surprised if there is a counterexample in the case where $\mathcal O(C)$ is a single point. I'd appreciate any proof, reference, or counterexample concerning either of these last two cases, or a reference indicating that this problem remains open.