= Solution
A <conformal automorphism of the upper half-plane> is a real <Möbius transformation>
$$
\phi(z)=\frac{az+b}{cz+d},\qquad a,b,c,d\in\mathbb R,
\quad ad-bc>0,
$$
with coefficients determined up to a common nonzero factor. The boundary interval $(\infty,1)$ is the oriented real boundary arc from infinity to $1$, namely the finite real interval $(-\infty,1)$. The map must take the left endpoint $-1$ to infinity and the right endpoint infinity to $1$, while fixing zero. These three boundary values determine the <Möbius transformation> uniquely:
$$
\boxed{\phi(z)=\frac{z}{1+z}.}
$$
Its determinant is positive, $\operatorname{Im}\phi(z)=\operatorname{Im}z/|1+z|^2>0$, and on $(-1,\infty)$ its derivative is $(1+x)^{-2}>0$. Its endpoint limits are $-\infty$ at $-1+$ and $1$ at $+\infty$, so it has exactly the required boundary extension.
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