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)?



Conformal isomorphism uniquely determined by boundary identification?
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.


Painleve's theorem holds for any rectifiable curve, and the proof is elementary, and uses nothing but Cauchy integral formula.
Rectifiability condition is too strong, and there was much research in the end of 20th century with the aim of understanding to which closed sets Painleve's theorem
really extends.


