Suppose $f:\mathbb{R} \rightarrow \mathbb{R}$ is an isometry of the reals. Prove that $f$ is a symmetry around a point if and only if $f$ reverses orientation of $\mathbb{R}$.
The orientation of $\mathbb{R}$ is the same as its order.
Part 1: The assumption is $f$ is a symmetry and we want to conclude it reverses the order.
Part 2: The assumption is $f$ is an isometry reversing the order and we want to conclude $f$ is a symmetry.
I'm not even sure how to begin this proof, much less how to do the rest of it. Any help would be appreciated.
Isometry of the reals means it preserves distances, so the distance from $x$ to $0$ is the same as from $f(x)$ to $f(0)$, or $|f(x)-f(0)|=|x-0|$. So $f(x)=f(0)\pm x$, and you have to pick $-$ if $f$ reverses the orientation of $\mathbb{R}$.
Now both parts get solved if you show that for any $a\in\mathbb{R}$ the map $f(x)=a-x$ is a symmetry around a point (hint: find which point stays fixed) and an isometry.