Is there an $\epsilon$- self isometry such that it is not an isometry? Let $f:(X, d)\to (X, d)$ be a a bijection on metric space $(X, d)$. 
For $\epsilon>0$, the bijection map $f$ is called $\epsilon$- self isometry, whenever $$\sup_{x,y\in X}|d(x, y)- d(f(x), f(y)|\leq \epsilon.$$ It is clear that if $f:(X, d)\to (X, d)$ is an isometry ( that is $d(f(x), f(y))=d(x, y)$), then it is an $\epsilon$- self isometry, for all $\epsilon>0$. 
Question. Let $\epsilon>0$ be given. Is there an $\epsilon$- self isometry such that it is not an isometry?
 A: Let $X$ be a metric space whose topology is nondiscrete, i.e. $X$ contains a non-isolated point $x$. Then for each $\epsilon>0$ there exists an $\epsilon$-isometry $f: X\to X$ which is not an isometry. To prove this, take a point $x\in X$ which is the limit of a sequence of pairwise distinct points $x_n$. Now, take $f(x)=x_n$, $f(y)=y$ for all $y\in X - \{x\}$. Then for all sufficiently large $n$, $f$ is an $\epsilon$-isometry but not an isometry.
Edit. Since you want bijective maps, take $f(x)=x_n, f(x_n)=x$ and $f(y)=y$ for all $y\notin \{x, x_n\}$. This map is discontinuous at $x$, hence, not an isometry. 
A: Consider $$X=\{ (x,0)\in \mathbb{E}^2| x\geq 1\} \cup \{ (x,1/x) \in
 \mathbb{E}^2| x\geq 1\} $$ which have a metric induced from the standard metric on
$\mathbb{E}^2$.
Define $f_n(x,0)=(x,0),\ f_n(x,1/x)=(x,1/x)$ where $x<n$ and
$f_n(x,0)=(x,1/x),\ f_n(x,1/x)=(x,0)$ where $x\geq
n$.
So $$ ||f_n(p)-f_n(q)| -|p-q|| \leq |f_n(p)-p| + |f_n(q) -q| \leq
 \frac{2}{n}$$
A: Yes, there is an $\epsilon$-isometry of $\mathbb{R}^n$ which is not an isometry. One idea is to develop a map which contracts $\mathbb{R}^n$ near the origin, but only barely moves other points. This is the approach I'll take here.
Consider a "radial contraction function" $$g(r) \equiv \begin{cases}1 &\text{if }r \geq 1 \\ r^2 & \text{if }r < 1\\\end{cases}$$
As a first pass, let's define a bijection $\widetilde f: \mathbb{R}^n\rightarrow \mathbb{R}^n$ which scales each vector by an amount $g(||x||)$; that is, it scales each point by an amount that depends on how far the point is from the origin:
$$\widetilde f (x) \equiv g(||x||)\cdot x $$
Note how this bijection behaves: points outside the unit ball are not moved at all, while points within the unit ball are moved slightly closer to the origin. Moreover, intuitively, the radius of that unit ball constrains how far points can move: we must have that for every point $x\in \mathbb{R}^n$,
$$|x - f(x)| < 2$$
since points outside the unit ball don't move, and points within the unit ball remain within the unit ball.
Hence we have for all $x$ and $y$:
\begin{align}
d(|x-y|, |f(x)-f(y)|)  &\leq  |x - f(x)| + |f(y)-y| & \text{triangle inequality}\\
&\leq 2 + 2 & \text{unit ball constraint}\\ 
& = 4
\end{align}
So even this crude function $\widetilde{f}$ is an $\epsilon$-isomorphism (for $\epsilon = 4$) while definitely not an isomorphism (since points in the unit ball are brought closer together).
We can make this trick work more generally, for any $\epsilon > 0$, by shrinking the radius of the ball where points are allowed to move: define
$$f(x) = g\left(\frac{4}{\epsilon}||x||\right) \cdot x.$$
Scaling the argument of $g$ has the effect of shrinking the radius of the ball to size $\epsilon/4$. Hence, by the same reasoning we have already employed, we find that for all points $x$ and $y$,
$$d(|x-y|, |f(x) - f(y)|) \leq 4\frac{\epsilon}{4} = \epsilon.$$
Q.E.D.
A: Fix $\epsilon > 0$. If $f:X \to X$ is a bijection satisying $d(x, f(x)) \leq \epsilon/2$ for all $x$, then $f$ is an $\epsilon$-self isometry by the triangle inequality: For all $x$ and $y$ in $X$,
\begin{gather*}
  d(x, y) \leq d(x, f(x)) + d(f(x), f(y)) + d(f(y), y) \leq d(f(x), f(y)) + \epsilon, \\
  d(f(x), f(y)) \leq d(x, f(x)) + d(x, y) + d(f(y), y) \leq d(x, y) + \epsilon,
\end{gather*}
from which it follows at once that
$$
\bigl|d(x, y) - d(f(x), f(y))\bigr| \leq \epsilon\quad\text{for all $x$, $y$ in $X$.}
$$
Bijections of this type are easy to find in Euclidean space (the space of interest according to the comments): Just swap a pair of points whose distance is at most $\epsilon/2$. For a continuous example, take an arbitrary continuous bijection that fixes every point outside some ball of radius $\epsilon/4$, or use Anthony's idea of a small-time flow of a bounded vector field.
A: These are plentiful; we can even demand the map be a homeomorphism
Let $(X,d)$ be your favourite space with finite-diameter $D$ and $f : X \to X$ your favourite homeomorphism that is not an isometry. Let $(X,\rho)$ be a new metric space where we define. . .  
$\displaystyle \rho(x,y) = \frac{\epsilon d(x,y)}{2D} \displaystyle$ 
Exercise: show $f$ is a homeomorphism but not an isometry on $(X,\rho)$.
We claim $f$ is an $\epsilon$ self isometry on $(X,\rho)$. This is because. . . 
$\sup|\rho(x, y)- \rho(f(x), f(y)| \le \sup \Big (|\rho(x, y)|+ |\rho(f(x), f(y)|\Big) < 2$diam$(X,\rho)$
Exercise: show $(X,\rho)$ has diameter $\epsilon/2$.
We conclude $\sup|\rho(x, y)- d(f(x), f(y)| \le \epsilon$
