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Let $(X, d) $ be a metric space and $f: X rightarrow X$ be a homeomorphism on $X$.

1) We say that $f$ is semi-equicontinuous if for every $epsilon > 0$ there exists $delta > 0$ such that $d(x, y) < delta $ implies $d(f^{n} (x), f^{n} (y)) < epsilon $ for all $x, yin X$ and all $n geq 0$. (This is also a definition of equicontinuity of continuous maps).

2) We say that $f$ is equicontinuous if for every $epsilon > 0$ there exists $delta > 0$ such that $d(x, y) < delta $ implies $d(f^{n} (x), f^{n} (y)) < epsilon $ for all $x, yin X$ and all $n in mathbb{Z}$.

3) We say that $f$ is an isometry if $d(f(x), f(y)) = d(x, y)$ for all $x, yin X$.

4) If $X= mathbb{R} $ and $0 < alpha < 1$, then $f(x) = alpha x$ is non-isometry semi equicontinuous homeomorphism but not equicontinuous.

Questions :

1) Can we find a non-isometry semi-equicontinuous but not equicontinuous homeomorphism on a compact metric space. A class of such maps on distinct compact metric spaces are needed, if exists.

2) If $f$ is an equicontinuous homeomorphism on a compact metric then $f$ must be an isometry?

2) A family $mathcal F$ of functions between metric spaces $(X,d)$ and $(X’,d’)$ is uniformly equicontinuous, if for every $epsilon>0$ there exists $delta>0$ such that $d(x,y)<delta$ implies $d’(f(x),f(y))<epsilon$ for all $x,yin X$ and all $gin mathcal F$. So a homeomorphism $f$ of a space $(X,d)$ is semi-equicontinuous iff a family $hat f^+={f^n:nge 0}$ is uniformly equicontinuous and is equicontinuous iff a family $hat f={f^n:ninBbb Z}$ is uniformly equicontinuous.

Uniform equicontinuity can be naturally defined for uniform spaces. Namely, a family $mathcal F$ of functions between uniform spaces $(X,mathcal U)$ and $(X’,mathcal U’)$ is uniformly equicontinuous, if for every $U’inmathcal U’$ there exists $Uinmathcal U$ such that $(ftimes f)(U)subset U’$ for all $ginmathcal F$.

It is easy to check that a family $mathcal F$ of functions between metric spaces $(X,d)$ and $(X’,d’)$ is uniformly equicontinuous iff $mathcal F$ is uniformly equicontinuous with respect to uniform spaces $(X,mathcal U_d)$ and $(X’,mathcal U_{d’})$, where $mathcal U_d$ and $mathcal U_{d’}$ are the uniformities induced by metrics $d$ and $d’$, respectively. That is a base of $mathcal U_d$ is ${U_n:ninBbb N}$, where $U_n={(x,y)in Xtimes X: d(x,y)<1/n}$ for each $ninBbb N$, and, similarly, a base of $mathcal U_{d}$ is ${U’_n:ninBbb N}$, where $U’_n={(x,y)in X’times X’: d’(x,y)<1/n}$ for each $ninBbb N$.

It is easy to check that a topology induced by the uniformity $mathcal U_d$ on the set $X$ is the same as a topology $tau_d$, induced by the metric $d$ on $X$.

At last we came to a key point of this answer. If $X$ is a compact space then there exists a unique uniformity $mathcal U$ on $X$, inducing the topology of space $X$, see, for instance, Theorem 8.3.13 from “General topology” by Engelking below. On the other hand, there can be many metrics on $X$ inducing its topology. We can construct them as follows. Let $(X,d)$ be any metric space and $h:Xto X$ be any continuous function. For each $x,yin X$ put $d’(x,y)=d(x,y)+|h(x)-h(y)|$. It is easy to see that a function $d’$ is a metric on $X$. Since $d’(x,y)ge d(x,y) $ for each $x,yin X$, a topology $tau_{d’}$ induced on $X$ by $d’$ is stronger than a topology $tau_d$ induced on $X$ by $d$. On the other hand, let $xin X$ be any point and $epsilon>0$ be any real number. Since the function $h$ is continuous at $x$, there exist $0<deltaepsilon/2$ such that $|h(x)-h(y)|<epsilon/2$ for each $yin X$ such that $d(x,y)<delta$. Then $d’(x,y)<epsilon$. Thus $tau_{d’}$ is weaker than $tau_d$, so $tau_{d’}=tau_d$.

We can conclude that if $X$ is a compact metrizable space and $f$ is a homeomorphism of $X$ then whether $X$ is equicontionuous does not depend on a metric inducing the topology of the space $X$, but
whether $X$ is an isometry depends on this metric and holds only in special cases.