Must an onto isometry map 'corner' point of a square to another 'corner' point of a square? Suppose that $T:(\mathbf{R}^2, \|\cdot\|_{\infty}) \rightarrow (\mathbf{R}^2, \|\cdot\|_{\infty})$ is an onto linear isometry, which leads it to bijective isometry with isometry inverse. 
The norm $\|\cdot\|_{\infty}$ is defined as 
$$\|(a,b)\|_{\infty} = \max\{ |a|,|b|\}.$$
Recall that a unit ball with the above norm is a square centered at origin with unit length.
We assume that both domain of codomain of $T$ do not contain $\ell^{\infty}(2).$
For every $(a,b) \in \mathbf{R}^2,$ denote 
$$T(a,b) = (c,d)$$

Question: For all $(a,b) \in \mathbf{R}^2$ with $|a| = |b| =1$, can we have $|c| = |d| = 1$ where $T(a,b) = (c,d)$?

In some sense, I am asking if I am given a 'corner' point on a square centered at origin with length $1,$ must the point be mapped to another 'corner' point of a square centered at origin with length $1$ under an onto isometry.
 A: A linear isometry $T$ preserves extreme points of the unit ball, i.e. $p$ is an extreme point of the unit ball if and only if $T(p)$ is one. Since the extreme points of the unit square are its corners, indeed a linear isometry must map corners of the unit square to corners of the unit square.
A: Note: I started writing this before saying I could assume linearity, but of course, it should hold for any isometry.
To answer what I think is the spirit of your question, yes, a surjective isometry from $(\mathbb{R}^2, \| \cdot \|_\infty)$ must map spheres to spheres and preserve their corners.
To see this, look at the metric line segment from the centre of the sphere out to the points on the sphere. In fact, I think it helps just to look at the set,
$$M(x, y) = \left\lbrace z \in \mathbb{R}^2 : \|x - z\| = \|z - y\| = \frac{\|x - y\|}{2}\right\rbrace.$$
Looking at it geometrically, it shouldn't be difficult to see that $M(x, y)$ is a singleton if and only if $x$ is on the corner of a sphere centred at $y$ (and vice-versa, of course). It's also easy to see, from the definition of isometry, that,
$$T(M(x, y)) = M(T(x), T(y)),$$
so $M(x, y)$ is a singleton if and only if $M(T(x), T(y))$ is a singleton. Thus, the "corner" relation is preserved.
