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Let $x_1 < x_2 < x_3 < x_4$ be real numbers. Let $\lambda$ be a positive real number larger than 1. How do we construct a conformal mapping on the extended complex plane that sends $x_1,x_2,x_3,x_4$ to $-\lambda, -1, 1, \lambda$ respectively?

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    Do you mean a Möbius transformation? – 2017-02-04
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    I think a possible answer could be a Möbius transform but not sure if a Möbius transform is the only possible answer – 2017-02-04

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If you want the conformal mapping to be defined on the Riemann sphere (I suppose this is what you mean by the extended complex plane?) then it has to be a Möbius transformation. Then note that any Möbius transformation has to preserve crossratios, i.e. writing:

$$ R(x_1,x_2;x_3,x_4) = \frac{(x_1-x_3)(x_2-x_4)}{(x_1-x_4)(x_2-x_3)} $$ then we have the auxiliary condition on $\lambda$ (up to permutations on the $x_i$'s): $$ R(x_1,x_2;x_3,x_4) = R(\lambda,1; -1,-\lambda) $$ In order to construct $M(z)=(az+b)/(cz+d)$ the easiest is perhaps to map $x_1,x_2,x_3$ first to $0,1,\infty$ and then to $\lambda,1,-1$, say. Find the value of $\lambda$ that verifies the auxiliary condition (then automatically $x_4$ will be mapped to -$\lambda$). For the first map you may take: $$ M(z) = \frac{(z-x_1)(x_2-x_3)}{(z-x_3)(x_2-x_1)}$$ For the second it should be: $$ N(z) = \frac{(\lambda-1)z+\lambda}{(1-\lambda)z-2} $$ then compose the two (but it will look quite horrible)

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    Thank you very much! I originally interpreted my own question to mean that $\lambda$ was to be fixed beforehand - and that I have realized is erroneous; indeed your solution is the correct interpretation, namely, that $\lambda$ is determined uniquely by $x_1,x_2,x_3,x_4$. The solution you provided is very clear. My appreciation! – 2017-02-05
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    You are welcome! – 2017-02-05