Equicontinuous homeomorphism and compact metric 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 $x, y\in X$ 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 $x, y\in X$ 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 $x, y\in X$. 
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?
 A: 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 $x,y\in X$ and all $g\in \mathcal F$. So a homeomorphism $f$ of a space $(X,d)$ is semi-equicontinuous iff a family $\hat f^+=\{f^n:n\ge 0\}$ is uniformly equicontinuous and is equicontinuous iff a family $\hat f=\{f^n:n\in\Bbb Z\}$ 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 $U’\in\mathcal U’$ there exists $U\in\mathcal U$ such that $(f\times f)(U)\subset U’$ for all $g\in\mathcal F$. 
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 $\{U_n:n\in\Bbb N\}$, where $U_n=\{(x,y)\in X\times X: d(x,y)<1/n\}$ for each $n\in\Bbb N$, and, similarly, a base of $\mathcal U_{d}$ is $\{U’_n:n\in\Bbb N\}$, where $U’_n=\{(x,y)\in X’\times X’: d’(x,y)<1/n\}$ for each $n\in\Bbb N$.
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 $h:X\to X$ be any continuous function. For each $x,y\in X$ 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 $x,y\in X$, 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 $x\in X$ be any point and $\epsilon>0$ be any real number. Since the function $h$ is continuous at $x$, there exist $0<\delta\epsilon/2$  such that $|h(x)-h(y)|<\epsilon/2$ for each $y\in X$ 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. 




