Let
$$ X=\{ \{x_n \}_{n=1} ^{ \infty } \mid {x_n } \to 0\ \},$$
where ${x_n }$ is a real sequence and put the metric $ d(x_n ,y_n)= \sup\limits_{n \in \Bbb {N}} |x_n -y_n|$ on $X$.
Let $X$ and $Y$ be metric spaces with metrics $d_X$ and $d_Y$. A map $f : X \to Y$ is called an isometry or distance preserving if for any $a,b \in X$ one has:
$${d_{Y}\left(f(a),f(b)\right)=d_{X}(a,b) }.$$
Now I have two questions:
1- find all isometry from $X$ to $X$.
2-find all isometry from $\Bbb{R}$ to $\Bbb{R}$ with the metric $d(a,b) = \max\{|a|,|b| \}$ on $\Bbb{R}^2$.
I find $f: X \to X$ , $ f(\{x_n \})=\{x_{P(n)} \} $ and $P: \Bbb{N} \to \Bbb{N}$ is a function that is one to one and onto (permutaion).
We know that an isometry is automatically injective and uniform continuous .for example $f(x)=x$ is trivial isometry. is group of isometry in above questions finite ? so give an example of two metric spaces that are homeomorphic but the group of isometry is not isomorphic.
This is not an anwser but it's too long for a comment.
-Your metric on $\mathbb{R}$ is not a metric, because $d(1,1) \neq 0$.
-Could you explain how you found your result about the isometries $X \rightarrow X$?
Note that the metric on $X$ can be induced by a norm on $X$, so every translation of $X$ is an isometry, $-id_X$ is an isometry, and you can find other types of isometries that do not come from permutations. (for instance adding a fixed value at a fixed place in a sequence: those are not surjective).
Finally, if you are looking for surjecive isometries in real normed vector spaces, by the Mazur-Ulam theorem, they are all affine, but even in that case it might not be easy to find all of them, just like it's not easy to determine a group of automorphisms in general.