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For linear fractional transformations, it suffices to specify where three points are mapped to. I am wondering if some analogue of this holds for general conformal mappings.

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No. For any finite collection of points $\zeta_1$, $\zeta_2$, ..., $\zeta_N$ in the disc of radius $1$, consider the map $$z \mapsto z+\epsilon \prod(z-\zeta_i).$$ This is a holomorphic map fixing all the $\zeta_i$ and, if $\epsilon$ is small enough, it is conformal.

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This depends on the group of conformal automorphisms that your region admits. E.g., the Riemann sphere admits a triply transitive group of conformal automorphisms, so you can prescribe the images of three arbitrary chosen points. If your region is the unit disc $D$ then the group of conformal automorphisms has "three real degrees of freedom". This means that you can prescribe, e.g., $f(0)\in D$ and $\arg(f'(0))$, but not $f(0)$ and $f({1\over2})$ independently.

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