Suppose $d:\mathbb{R}\times\mathbb{R}\rightarrow\mathbb{R}$ is a map such that there exists a bijective function $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $$ d(x, y) = |f(x) - f(y)|. $$ Let us call $d(x, y)$ the distance between $x$ and $y$. Suppose the distance between points is invariant under reflections of the real line about points. That is, for all $r\in\mathbb{R}$ we have $d(x, y) = d(2r-x, 2r-y)$ (the map $x\mapsto 2r-x$ is a reflection about $r$).
Can we deduce that for some positive $\alpha$ we have $d(x, y) = \alpha|x-y|$ for all $x, y\in\mathbb{R}$?
Progress So Far
Without loss of generality, we may assume $f(0) = 0$. It helps to realize that a translation of the real line is a composition of two reflections, so our distance function is invariant under translations.
I've been able to show $f(p/q) = \delta_{1} p/q$ for all rational $p/q$ where $\delta_{1}$ is some nonzero constant. Similarly, if $\varepsilon$ is an irrational, then $f(\varepsilon p/q) = \delta_{\varepsilon}p/q$ for all rational $p/q$ where $\delta_{\varepsilon}$ is some nonzero constant.
I am left wondering if translation and reflection invariance in the way I posited above can allow me to deduce $\delta_{\varepsilon} = \delta_{1}\varepsilon$. Remember that $f$ is bijective but not necessarily continuous.
There exist solutions to the functional equation $f(x+y) = f(x) + f(y)$ which are not $\mathbb{R}$-linear. (This is called Cauchy's functional equation.) As you prove, though, they are $\mathbb{Q}$-linear.
Any such $f$ gives rise to a $d$ invariant under reflections.