Now find all isometry from $\Bbb{R^2}$ to $\Bbb{R^2}$ with the norm $\|(a,b) \|= \max\{|a|,|b| \}$ on $\Bbb{R}^2$. in my book definition of isometry is: 
Let $X$ and $Y$ be normed spaces with norms  $\left\|\cdot\right\|_1$ and  $\left\|\cdot\right\|_2$. A map $f : X \to Y$ is called an isometry if for any $a,b \in X$ one has:
$$\left\|f(a)-f(b)\right\|_1=\|a-b\|_2.$$
(functional analysis by Peter D.Lax)
Now find all isometry from $\Bbb{R^2}$ to $\Bbb{R^2}$ with the norm $\|(a,b) \|=  \max\{|a|,|b| \}$ on $\Bbb{R}^2$.
I find some of them in this space :
\begin{align*}f(x,y) &= (x,y)\\
f(x,y) &= (-x,y)\\
f(x,y) &= (x,-y)\\
f(x,y) &= (y,x)\\
f(x,y) &= (-y,x)\\
f(x,y) &= (y,-x)\\
f(x,y) &= (y+c,x+c)\qquad (c \in\mathbb{R})\\
f(x,y) &= (x+c,y+c). 
\end{align*}
We know that an isometry is automatically injective and uniform continuous. So, give an example of two normed spaces that are NOT isometric but the group of isometry is isomorphic.
 A: Translation by a vector is always an isometry, in any normed space. So we can stick to the isometries that fix $0$. Those leave the closed unit ball invariant, and most notably, leave the set of its extreme points invariant. In case of $\mathbb R^2$ with the maximum norm, the extreme points of the closed unit ball are $(\pm 1,\pm 1)$. So there are four choices for the image of $(1,1)$ under an isometry, and after that there are two choices for its neighbor $(1,-1)$. Thus, there are eight isometries; your list is missing $(x,y)\mapsto (-x,-y)$ and $(x,y)\mapsto (-y,-x)$. 

give an example of two normed spaces that are NOT isometric but the group of isometry is isomorphic.

For a generic normed space over reals (just draw a random central-symmetric convex shape and call it the unit ball) the group of isometries is cyclic of order $2$: it consists of the identity and $\vec x\mapsto -\vec x$. This is one approach. Another one is to consider $\mathbb{R}^2$ with the norm $\|(x,y)\|_p = (|x|^p+|y|^p)^{1/p}$; when $p\in (1,\infty)\setminus \{2\}$, the isometry group is the same as of the aforementioned space with the maximum norm; but the spaces are not isometric since the spaces with $1<p<\infty$ are strictly convex.
