# Conformal isomorphism uniquely determined by boundary identification?

Let $$\Gamma$$ be a smooth Jordan arc, and let $$\Phi \colon \hat{\mathbb C} \backslash \overline{\mathbb D} \to \hat{\mathbb C} \backslash \Gamma$$ be a conformal isomorphism that fixes the point at $$\infty$$. Then $$\Phi$$ extends continuously onto $$\partial \mathbb D$$ because $$\Gamma$$ is smooth (Caratheodory), and for every $$z \in \Gamma$$ except the two endpoints, $$\Phi^{-1}(z)$$ consists of two points on $$S^1$$.

Now if I have another smooth Jordan arc $$\Gamma_0$$, and a conformal isomorphism $$\Phi_0 \colon \hat{\mathbb C} \backslash \overline{\mathbb D} \to \hat{\mathbb C} \backslash \Gamma_0$$ that fixes $$\infty$$ such that $$\Phi_0(\Phi^{-1}(z))$$ is a singleton for every $$z \in \Gamma$$, can I necessarily conclude that $$\Gamma$$ and $$\Gamma_0$$ are the same curve (modulo affine transformation)?

For smooth curves, the answer is yes. Proof: $$\Phi_0\circ\Phi_1^{-1}$$ is a conformal map of $$C\backslash\Gamma_0$$ onto $$C\backslash\Gamma$$, and your condition implies that this conformal map extends continuously to the boundary. So we have a continuous map (in fact a homeomorphism) of the Riemann sphere, which is conformal in the complement of a smooth curve. By a theorem of Painleve, such a map is conformal on the whole sphere, so it is linear-fractional, and since it fixes $$\infty$$ it is affine.