Isometry, compact space, there exists only one isometry other than identity I'm solving a problem concerning isometry in a compact metric space $([0,1],d)$, the metric $d$ is consistent with natural topology of $[0,1]$. 
Here is the problem: Prove that in the space defined above there exists only one isometry other than identity.
Is it true that if $f$ is the isometry, then the sequence $(f^n(x)), \ n \in \mathbb{N}$ is strictly monotone and if so, how to prove it?
 A: Note that $0$ and $1$ play a special role in $X=[0,1]$: They are the only points $x\in X$ such that $X\setminus \{x\}$ is a connected space. Such properties are respected by isometries (which are especially homeomorphisms), hence an isometry $f$ can either leave $0$ and $1$ fixed or swap them.
It is sufficient to show that an isometry with $f(0)=0$ and $f(1)=1$ is necessarily the identity. (Why? If we have shown this fact and there are two non-identity isometries $g,h$, then these must both be swapping, hence $g\circ h$ and $g\circ g$ are non-swapping isometries and then $g\circ h=\operatorname{id}=g\circ g$ implies $h=g$).
Now let $f$ be an isometry with $f(0)=0$ and $f(1)=1$.
Assuming $f\ne\operatorname{id}$, the set $\{x\in X\mid f(x)\ne x\}$ is open and, as it does not contain $0$ or $1$, is also open as subset in $\mathbb R$, hence is the union of disjoint open intervals. Let $(a,b)$ be one of those intervals, i.e. $f(a)=a$, $f(b)=b$, but $f(x)\ne x$ for $a<x<b$.
Then either $f(x)<x$ for all $x\in(a,b)$ or $f(x)>x$ for all $x\in(a,b)$ (because $x\mapsto \frac{f(x)-x}{|f(x)-x|}$ is a continuous map from the connected set $(a,b)$ to $\{-1,1\}$).
We consider only the case that $f(x)<x$ for all $x\in(a,b)$, the other case can be treated similarly.
Let $t\in[a,b)$ be a point with $f(t)\le a$. Then by the IVT, there is $\xi\in[t,b)$ with $f(\xi)=a$. As $f$ is a $d$-isometry, we conclude $d(\xi,a)=d(f(\xi),f(a))=d(a,a)=0$, hence $\xi=a$ and finally $t=a$ from $a\le t\le \xi$.
In other words,  $a<f(t)<t$ for all $t\in(a,b)$.
Then for $x\in(a,b)$, the sequence $(f^n(x))_{n\in\mathbb N}$ obtained by iterating is strictly decreasing and bounded from below by $a$, hence converges to some limit $y\in[a,b)$. As $f$ is continuous, we have $f(y)=y$, i.e. $y=a$.
Let $r=d(x,a)$. Then we have just shown that in any open neighbourhood $U$ of $a$, there are points $z=f^n(x)$ with $d(a,z)=r$. If we let $U=\{z\in X\mid d(a,z)<r\}$, we get a contradiction.
Therefore the assumption $f\ne\operatorname{id}$ is false.

You may notice that I used that $(f^n(x))_{n\in\mathbb N}$ is a strictly decreasing sequence in my proof (and would be strictly increasing in the case where $f(x)>x$ for all $x\in(a,b)$).
However, this appears while discussing a hypothetical isometry that is not the identity and does not swap the end points.
It is not true that $(f^n(x))_{n\in\mathbb N}$ is strictly monotone in any isometry of $[0,1]$. In fact, the sequence is constant if $f$ is the identity and if $f$ is the other possible isometry, the sequence e.g. for $x=0$ runs $0,1,0,1,\ldots$, i.e. is not monotone.

We have shown that apart from the identity there is at most one other isometry and that such an isometry must swap $0$ and $1$.
But it is not necessarily the case that such a swapping isometry actually exists.
Consider the embedding $\iota\colon [0,1]\to\mathbb R^2$, $t\mapsto (\frac1{50}t,2t^2-t)$ (sketch it!). Then $d(x,y):=\lVert \iota(x)-\iota(y)\rVert$ is a metric that induces the standard topology on $[0,1]$.
As $d(0,\frac14)>\left|2\cdot \frac1{4^2}-\frac14\right|=\frac18$, there exists some $x\in(0,\frac14)$ with $d(0,x)=\frac1{100}$. Also, $d(0,\frac12)=\frac1{100}$, so there are at least two (in fact, three) points at distance $\frac1{100}$ from $0$. On the other hand, there is only one point at distance $\frac1{100}$ from $1$. Therefore, a $d$-isometry cannot interchange $0$ and $1$.
