Find all functions $f : \mathbb{R} \rightarrow \mathbb{R}$ such that for all $I$ open bounded interval it follows $f(I)$ is also an open bounded interval of same length as $I$.
It's easy to see $f(x)=\pm x + c, c \in \mathbb{R}$ are solutions, but I cannot prove/disprove these are all.
Intuitively, it seems that $|f(x) - f(y)|=|x - y|$ should stand for all $x, y$, following the same length condition.
Hint: $f$ must be monotone, so $(f(x)-f(y))/(x-y)$ should be the same for all $y \ne x$.
EDIT: To show $f$ is monotone: if $a < b < c$, $f((a,c)) = f((a,b)) \cup f((b,c)) \cup \{f(b)\}$. Now the length of $f((a,c))$ is the length of $(a,c)$, which is the sum of the lengths of $(a,b)$ and $(b,c)$, and thus the sum of the lengths of $f((a,b))$ and $f((b,c))$. The only way the union of two open intervals (plus a point) can have length equal to the sum of the lengths of the intervals is that those intervals are disjoint.
Now if $f$ was not monotone, there would be some $x < y < z$ such that $f(y)$ is not between $f(x)$ and $f(z)$: either $f(y) \ge \max(f(x), f(z))$ or $f(y) \le \min(f(x), f(z))$. Suppose $f(x) \le f(z) \le f(y)$ (the other cases will be similar). Then taking $a < x < y < b < z < c$, $f((a,b))$ and $f((b,c))$ both include $f(z)$ and are thus not disjoint, contradiction.