Consider a bijection between two affine spaces of the same dimension $n$ (let's assume $n\ge 1$ to avoid trivialities) which sends any three collinear points into collinear points.
Must such a map be affine?
I proved this with additional assumption that the affine ratio of these three points is preserved, but it seems much harder in the general case.
Yes, it can be proven by induction and the case for two dimensions seem most technical.
First we note that a bijective collineation will map parallell hyperplanes to parallell hyperplanes. Then we note that since composing with an affine mapping will not alter anything. This means that we can assume that the mapping maps hyperplanes parallell to the $x=0$ hyper plane to hyperplanes parallell to $x=0$ and that it maps origin to origin.
For two dimension this means that we can reduce the problem to the mapping $(x,y)\to f(x), g(x)$. Since we can scale axes we can assume that $(1,1)\to f(1), g(1) = (1,1)$ so we have by collinearity that $(x,x)\to f(x), g(x)$ being on the line $x-y=0$ so $f(x)=g(x)$. Then by considering lines through $(0,1)$ and $(x, 0)$ together with the intermediate point $(x/(x+1), x/(x+1)$ you can conclude that $f(x) = x$.
Then one can use induction for the rest as the mapping can after composition be written as $(x, u)\to f(x), L(u)$ where $L(u)$ is a linear mapping of the hyperplanes and $f(x)$ is some mapping and since the line through origin and $(1,1,\cdots,1)$ have to map to a line we see that $f(x)$ must be linear.
For the cases you omit the situation is obvious, but for the record I mention that for dimension $0$ all mappings are affine so it's true there, but for dimension $1$ it's not since any mapping are preserving collinearity, but not all bijective mappings are not affine.