Interval preserving transformations are linear in special relativity

In hindsight, here is a short proof.

The metric $g_{\mu\nu}$ is the flat constant metric $\eta_{\mu\nu}$ in both coordinate systems. Therefore, the corresponding (uniquely defined) Levi-Civita Christoffel symbols

$$ \Gamma^{\lambda}_{\mu\nu}~=~0$$

are zero in both coordinate systems. It is well-known that the Christoffel symbol does not transform as a tensor under a local coordinate transformation $x^{\mu} \to y^{\rho}=y^{\rho}(x)$, but rather with an inhomogeneous term, which is built from the second derivative of the coordinate transformation,

$$\frac{\partial y^{\tau}}{\partial x^{\lambda}} \Gamma^{(x)\lambda}_{\mu\nu} ~=~\frac{\partial y^{\rho}}{\partial x^{\mu}}\, \frac{\partial y^{\sigma}}{\partial x^{\nu}}\, \Gamma^{(y)\tau}_{\rho\sigma}+ \frac{\partial^2 y^{\tau}}{\partial x^{\mu} \partial x^{\nu}}. $$

Hence all the second derivatives are zero,

$$ \frac{\partial^2 y^{\tau}}{\partial x^{\mu} \partial x^{\nu}}~=~0, $$

i.e. the transformation $x^{\mu} \to y^{\rho}=y^{\rho}(x)$ is affine.

I had the feeling that a direct proof would be possible using only the relation $\eta _{ij}\frac{\partial y^i}{\partial x^p}\frac{\partial y^j}{\partial x^q}=\eta _{pq}$, assuming simple smoothness properties of the transformation and then using some algebra maneuvers. I found the following lovely argument in the book Gravitation and Cosmology by Steven Weinberg.

We start from the relation

$$\eta _{ij}\frac{\partial y^i}{\partial x^p}\frac{\partial y^j}{\partial x^q}=\eta _{pq}$$

Differentiating with respect to $x^k$ we obtain

$$\eta _{ij}\frac{\partial ^2y^i}{\partial x^p\partial x^k}\frac{\partial y^j}{\partial x^q}+\eta _{ij}\frac{\partial y^i}{\partial x^p}\frac{\partial ^2y^j}{\partial x^q\partial x^k}=0$$

We add to this the same equation with $p$ and $k$ interchanged, and subtract the same with $q$ and $k$ interchanged; that is,

$$\eta _{ij}\left(\frac{\partial ^2y^i}{\partial x^p\partial x^k}\frac{\partial y^j}{\partial x^q}+\frac{\partial y^i}{\partial x^p}\frac{\partial ^2y^j}{\partial x^q\partial x^k}+\frac{\partial ^2y^i}{\partial x^k\partial x^p}\frac{\partial y^j}{\partial x^q}+\frac{\partial y^i}{\partial x^k}\frac{\partial ^2y^j}{\partial x^q\partial x^p}-\frac{\partial ^2y^i}{\partial x^p\partial x^q}\frac{\partial y^j}{\partial x^k}-\frac{\partial y^i}{\partial x^p}\frac{\partial ^2y^j}{\partial x^k\partial x^q}\right)=0$$

This simplifies to

$$2\eta _{ij}\frac{\partial ^2y^i}{\partial x^p\partial x^k}\frac{\partial y^j}{\partial x^q}=0$$

Since the tensors $\frac{\partial y^i}{\partial x^j}$ and $\eta _{ij}$ are invertible, this implies that

$$\frac{\partial ^2y^i}{\partial x^p\partial x^k}=0$$

Here I just want to mention that there exists a direct proof in $1+1$ dimensions using elementary arguments. Let the two coordinate patches $U_x$ and $U_y$ (which are, say, both convex sets in $\mathbb{R}^2$, containing the origin) have light-cone coordinates $x^{\pm}$ and $y^{\pm}$, respectively. The metric reads

$$ dy^{+}dy^{-} ~=~ dx^{+}dx^{-}. $$

This leads to three PDE's

$$ \frac{\partial y^{+}}{\partial x^{+}} \frac{\partial y^{-}}{\partial x^{+}} ~=~0 \qquad \qquad\Leftrightarrow\qquad \qquad\frac{\partial y^{+}}{\partial x^{+}}~=~0 \qquad\mathrm{or}\qquad \frac{\partial y^{-}}{\partial x^{+}} ~=~0 ,$$ $$ \frac{\partial y^{+}}{\partial x^{-}} \frac{\partial y^{-}}{\partial x^{-}} ~=~0\qquad \qquad\Leftrightarrow \qquad\qquad\frac{\partial y^{+}}{\partial x^{-}}~=~0 \qquad\mathrm{or}\qquad \frac{\partial y^{-}}{\partial x^{-}} ~=~0,$$ $$ \frac{\partial y^{+}}{\partial x^{+}} \frac{\partial y^{-}}{\partial x^{-}} +\frac{\partial y^{+}}{\partial x^{-}} \frac{\partial y^{-}}{\partial x^{+}} ~=~1.$$

Since $\det \frac{\partial y}{\partial x}\neq 0$, there are really only two possibilities. Either

$$\frac{\partial y^{-}}{\partial x^{+}}~=~0 ~=~\frac{\partial y^{+}}{\partial x^{-}},$$

or

$$\frac{\partial y^{+}}{\partial x^{+}}~=~0 ~=~\frac{\partial y^{-}}{\partial x^{-}}.$$

By possibly relabeling $x^{+} \leftrightarrow x^{-}$, we may assume the former. So

$$y^{+}~=~f^{+}(x^{+})\qquad \mathrm{and} \qquad y^{-}~=~f^{-}(x^{-}).$$

From the third PDE, we conclude that

$$ \frac{\partial f^{+}}{\partial x^{+}}\frac{\partial f^{-}}{\partial x^{-}} ~=~1. $$

By separation of variables, this is only possible if $\frac{\partial f^{\pm}}{\partial x^{\pm}}$ is independent of $x^{\pm}$. It follows that $y^{\pm}~=~f^{\pm}(x^{\pm})$ are affine functions. Q.E.D.