This question is related to a question I posted earlier today (see here, this metric gives me nightmares), but I think it is easier to answer this one. Let $m\in\mathbb{N}_{\geq2}$ and consider the map $f\colon[0,1)\to[0,1)$ defined by $f(x):=mx\mod 1$. Also consider the metric $d(x,y):=\min\{|x-y|,1-|x-y|\}$ on $[0,1)$. I think that the following statement is true, but I don't know how to prove it:
Suppose that $d(x,y)\leq\frac{1}{2m}$. Then $$d(f^{k}(x),f^{k}(y))=m^kd(x,y)$$ for all $0\leq k\leq m$.
I think that I know how to prove the statement for the cases $|x-y|\leq\frac{1}{2}$: In this case we have $d(x,y)=|x-y|$, so $$d(f^{k}(x),f^{k}(y))=|kx-ky\mod1|\overset{?}{=}k|x-y|.$$ However, I'm not sure about the step with the questionmark. For the case $|x-y|\geq\frac{1}{2}$, I have no idea. Any suggestions are greatly appreciated.
EDIT: I also know that this metric can be viewed as the metric on the unit circle that measures the shortest arc length between two points (divided by $2\pi$). However, this only gives me some intuition.
I think one can reduce your question to the case $|x - y| \leq 1/2$ as follows. If $|x - y| > 1/2$ then there exists an isometry of $[0,1)$, namely $h: z \mapsto z - y \text{ mod } 1$, mapping $y$ to $0$. But then, $|h(x) - h(y)| \leq 1/2$ (otherwise swap $x$ and $y$), so we can apply your result to $h(x)$ and $h(y)$.
Edit: Your original statement had a mistake. A complete solution to a corrected version could be outlined as follows. First, apply an isometry to $[0, 1)$ such that $x = 0$ and $y \leq 1/(2m)$ (thinking of $S^1$ helps). Then compute $$d(f(x), f(y)) = md(x,y).$$ Similarly, $$d(f^2(x), f^2(y)) = md(f(x), f(y)) = m^2d(x, y).$$ Proceeding similarly, $$d(f^k(x), f^k(y)) = md(f^{k-1}(x), f^{k-1}(y)) = \dots = m^kd(x,y).$$