If an isometry $f$ of a length space $X$ fixes a ball $B(x,r)$ for some $x$ and $r>0$, is it true that $f$ is the identity map?

Question

Let $(X,d,L)$ be a length space, that is, $d = \inf L$, where the inf is taken over all curves with fixed endpoints.

In $\mathbb{R}^2$ we have a results that says: if an isometry fixes 3 points, then it is the identity map. I was wondering if such result holds for $X$ changing "3 points" by an open ball $B(x,r)$, $r>0$. Note that since $X$ is a length space, then $X$ doesn't have discrete points, which would lead to a trivial counter-example.

EDIT: By "fixing" I actually mean being the identity in the ball. So the question is:

If $F : X \to X$ is an isometry of $X$ and there is a ball $B$ such that $F|_B = Id|_B$, then is it true that $F = Id$, where $Id :X \to X$ is the identity map?

Answer 1

Take $X$ to be the union of two coordinate axes in the Euclidean plane, $X=\{(x,y): xy=0\}$, with the induced path-metric from the plane. Now, consider $f(x,0)=(x,0)$ and $f(0,y)=(0,-y)$, $f: X\to X$. I will leave it to you to verify that $f$ is an isometry fixing metric balls of arbitrarily large radii.