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An order automorphism of an order-dense subset $D$ of the real line extends uniquely by $\tilde{\phi }(x)=\sup \{\phi (d):d\in D,d<x\}$. Order completeness supplies the extension, and the lower cuts in $D$ give uniqueness.

Use the order completeness of $\mathbb{R}$ to defineThe set in this supremum is nonempty and bounded above: choose $d_0,d_1\in D$ with $d_0<x<d_1$, using that $D$ is an order-dense subset. For $x\in D$, all its displayed values lie below $\phi (x)$. If $y<\phi (x)$, choose $e\in D$ with $y<e<\phi (x)$; then $d={\phi }^{-1}(e)<x$. Thus the supremum equals $\phi (x)$.

The extension is strictly increasing. If $x<y$, choose $d,e\in D$ with $x<d<e<y$; thenApply the same construction to ${\phi }^{-1}$, obtaining $\psi$. For $d<x$, choose $e\in D$ with $d<e<x$; this gives $\phi (d)<\tilde{\phi }(x)$. For $d>x$, a point of $D$ between $x$ and $d$ similarly gives $\phi (d)>\tilde{\phi }(x)$. HenceTaking inverse images and a supremum shows $\psi (\tilde{\phi }(x))=x$. The symmetric argument gives $\tilde{\phi }(\psi (y))=y$, so the extension is an order automorphism.

For uniqueness, any increasing extension $h$ has the same strict lower cut in $D$ at its value $h(x)$. Two distinct real numbers have different cuts in an order-dense subset, so $h(x)=\tilde{\phi }(x)$. The extension exists and is unique.