We know there are metric spaces with no non-trivial (global) isometries. For example, consider the Euclidean plane $(\mathbb{R}^2, d)$. Let $X \subset \mathbb{R}^2$ be the set $\{(0,0), (1,0),(0,2)\}$. Now consider the metric space $(X, d_X)$ where $d_X$ is the metric induced from $d$. One can easily verify that the only isometry $f: X \rightarrow X$ is the identity map.
My question is whether there are also metric spaces with no non-trivial "local isometries" of a certain kind. Here's what I mean by that...
Definition. We say a metric space $(X,d)$ has has a (non-trivial) local isometry if there are distinct points $p,q \in X$ with respective neighborhoods $O_p$ and $O_q$ (in the induced topology) and an isometry $f: O_p \rightarrow O_q$ such that $f(p)=q$.
There are cheap examples of metric spaces with no non-trivial local isometries. Just consider any metric space $(X,d)$ where $X$ contains only one element. My question is the following...
Question. Is there a a metric space with at least two elements which fails to have a non-trivial local isometry?