TITLE: Metric space has at most one isometry other than identity
QUESTION [7 upvotes]: Could you help me with this problem?
Let $d$ be a metric on $[0,1]$ consistent with the standard topology. Prove that the metric space: $([0,1], d)$ has at most one isometry (except for identity).
I would really appreciate your help.
Thank you.

REPLY [8 votes]: Notice that it is enough to show that an isometry which fixes the endpoints is the identity.
Suppose then $f:[0,1]\to[0,1]$ is an isometry such that $f(0)=0$, and let $x\in[0,1]$. Notice that $f$ is strictly increaing —because it is continuous for the usual topology and injective.
It is easy to check that the sequence $(f^n(x))_{n\geq1}$ is monotone, so that it converges with respect to the usual topology of $[0,1]$. The hypothesis then implies that it also converges with respect to $d$ and, in particular, it satisfies the condition of Cauchy. Since $f$ is an isometry, $d(f^n(x),f^{n+1}(x))$ does not depend on $n$ because $f$ is an isometry, and then the Cauchy condition implies that this distance is zero: in particular, $d(x,f(x))=0$, that is $x=f(x)$.