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.
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 $(x,y)\mapsto (-y,-x)$.
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.