Let $E$ be a real vector space, and $q$ be a quadratic form on $E$ of signature $(-1,1,\cdots,1)$. Let us call time vectors the vectors $v$ such that $q(v) < 0$. The set of time vectors has two connected components. Let $C^+$ be one of them. It is a convex cone, and gives an order on $E$ given by: for all $v,w \in E$, $v \leq w$ if $w -v \in C^+$.
Every affine bijection of $E$ whose linear part is the composition of a homothety of positive coefficient and an isometry of $q$ is order-preserving.
Is the converse true?
That is, let $f : E \rightarrow E$ be a bijection that is order-preserving. Is $f$ an affine map of the above type? If not, what if we also assume that $f^{-1}$ is order-preserving too?
It looks like from a theorem of Malament (click here) and a theorem of Hawking cited in the paper, assuming mild continuity conditions, such a map should be a diffeomorphism whose differential at any point is the product of a homothety and a linear map preserving the Lorentz form.
Consider the form $(x,y) \mapsto x y$ on $\mathbb{R}^2$, of signature $(1,-1)$. The causal order on $\mathbb{R}^2$ will be the product order.
Any product of isomorphisms of the usual order on $\mathbb{R}$ will be an iso of the product order. So any map $(x,y)\mapsto (f(x), g(y))$, where $f$, $g$ are increasing (homeomorphsisms of $\mathbb{R}$).
Note that the automorphisms of the form $x y$ are $(x,y) \mapsto ( t x , \frac{1}{t} y)$.
In this case it does not seem to imply the smoothness of the transform. Maybe in higher dimensions?
$\bf{Added}$ The answer is correct for Lorentz spaces of dimension $\ge 3$, cf the result of E.C. Zeeman, J. Math. Phys. 5, 490 (1964)