Is there a linear isometry between $(\Bbb R^N, \left \| \cdot \right \|_{1} )$ and $(\Bbb R^N, \left \| \cdot \right \|_{\infty} )$? An apology if I did not specify it well. I want to find a function $\varphi: \mathbb{R}^{N}\rightarrow\mathbb{R}^{N}, x \rightarrow \varphi \left(x \right)$ that is bijective , linear and preserving the norm, ie $\forall x \in \mathbb{R}^{N}, \left \| x \right \|_{1} = \left \| \varphi \left ( x \right) \right \|_{\infty}.$
I have dealt with some functions but I can not find a convenient one. 
I'm sorry, I mean isomorphic as topological spaces with the metrics there, but the isomorphism does not come with the identity, because it does not preserve the norm.
Thanks for the help.
 A: As @Batomintovski states, $(\mathbb{R}^n, \|\cdot\|_1)$ and $(\mathbb{R}^n, \|\cdot\|_\infty)$ are isometrically isomorphic if and only if $n \in \{1,2\}$.
For $n = 1$ the spaces are the same.
For $n=2$ check that the isometrical isomorphism $T : (\mathbb{R}^n, \|\cdot\|_\infty) \to (\mathbb{R}^n, \|\cdot\|_1)$ is given by
$$T(x,y) = \frac12 (x+y, x-y)$$
Indeed $$\|T(x,y)\|_1 =  \frac{|x-y| + |x+y|}{2} = \max\{|x|,|y| \} = \|(x,y)\|_\infty$$
Bijectivity follows from injectivity because of equality of dimensions.
For $n \ge 3$ count the number of extreme points on the unit ball in the two spaces.
For $(\mathbb{R}^n, \|\cdot\|_\infty)$ the extreme points are of the form $(\varepsilon_1, \ldots, \varepsilon_n)$ with $\varepsilon_i \in \{-1,1\}$. Hence there are $2^n$ extreme points.
For $(\mathbb{R}^n, \|\cdot\|_1)$ the extreme points are 
$$(\pm 1, 0, 0,\ldots, 0), (0,\pm 1,0, \ldots, 0), \ldots, (0,\ldots, 0, \pm 1)$$
so there are $2n$ of them.
For $n \ge 3$ we have $2^n \ne 2n$ so the spaces cannot be isometrically isomorphic.
