Metric space has at most one isometry other than identity 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.
 A: 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 $x=f(x)$.
A: Hint: Continuity is a topological property.
Spoiler: The following is the shape of a proof.


*

*Isometries are continuous functions.

*Continuity is a topological property.

*Isometries are bijections.

*Continuous bijections $([0,1],d_\textrm{Eucl.}) \to ([0,1],d_\textrm{Eucl.})$ are strictly increasing or strictly decreasing.

*A strictly increasing bijective function $[0,1]\to[0,1]$ which is an isometry w.r.t. the metric $d$ is unique.

*Similarly for a strictly decreasing bijective function.


Edit: For point 5, suppose $f,g$ both have this property. Then consider $h=f^{-1}\circ g$. This is a continuous bijection (check), so is strictly increasing/decreasing. Suppose $h(x)\neq x$. Then $d(h^n(x),h^{n-1}(x)) = d(h(x),x)\neq 0$. But let $n\to\infty$; by monotonicity, $h^n(x)\to x_0$, and by continuity, $d(h^n(x),h^{n-1}(x)) \to 0$. Contradiction.
