Banach space $\mathbb{C}^2$ with different norm Let $(\mathbb{C}^2,\|\cdot\|)$ be the Banach space over $\mathbb{C}$ with $\|(z_1,z_2)\|=\max\{|z_1|,|z_2|\}$ for any $(z_1,z_2)\in\mathbb{C}^2$. Let $w_1=(1,1)$ and $w_2=(1,0)$. Then show that there is NO linear isometry on $\mathbb{C}^2$ which maps $w_1$ to $w_2$.
The above problem looks very simple, and it may have a totally trivial solution but I am unable to prove it.
My Attempt :
It is immediate that $\|\cdot\|$ does not satisfy the parallelogram law on $\mathbb{C}^2$. So there does not exist any inner product on $\mathbb{C}^2$ which induces $\|\cdot\|$. I also noticed that the set $\{w_1,w_2\}$ is a basis of $\mathbb{C}^2$ over $\mathbb{C}$. If there exists a linear isometry $T$ on $\mathbb{C}^2$ such that $T(w_1)=w_2$ then $T$ is an isometric isomorphism.
Now I am not able to proceed further and obtain contradiction.
 A: General remark: by Theorem by Mazur-Ulam, every isometry (onto; but not necessarily linear) between two Banach spaces preserves the arithmetic mean of two points, i.e
$$ J\left(\frac{a+b}2\right)\ =
\ \frac{J(a)+J(b)}2 $$
Actually, by the same theorem, such $\ J\ $
must be linear when $\ J(0)=0.$

Notation: $\ \pi_1(z)\ $ and $\ \pi_2(z)\ \in \mathbb C\ $ are the two coordinates of point $\ z\in\mathbb C^2.$

THEOREM: there does not exist a linear isometry $\ J:\mathbb C^2\to \mathbb C^2\ $ such that $\ J(w_2)\ =\ w_1. $
PROOF: Let
$$ S\ :=\ \{w\in\mathbb C^2: \|w\|=1\} $$
be the unit sphere around the origin. Of course $\ w_k\in S\ $ both for $\ k=1\ $ and $\ k=2.$
Linear isometries map isometrically $\ S\ $ onto $\ S.\ $
Point $\ w_2\ $ is the central point between the following two points of $\ S,\ $ namely $\ w_1\ $ and $\,\ z:=2\cdot w_2-w_1\in S,\ $ i.e.
$$ w_2\ :=\ \frac{w_1+z}2 $$
where: $\ w_1-w_2\ $ and $\,\ z-w_2\ \in\ S,\ $ and $\ \frac{z-w_1}2\in S\ $ (since
$\ \frac{z-w_1}2\ =\ w_2-w_1\ $).
This metric situation of $\ w_2\ $ is impossible for $\ w_1.\ $
Indeed, let $\ u\ $ and $\ v\in S\ $ be such that all three hold: $\ u-w_1\in S\ $ and $\ v-w_1\ \in\ S,\ $ and $\frac{v-u}2\in S. $
Then
$$ \pi_k\left(\frac{v-u}2\right)\ \le\ \frac{\sqrt{3}}2\ <\ 1 $$
for both $\ k\ $ hence
$$ \left\|\frac{v-u}2\right\|\ \le\ \frac{\sqrt{3}}2\ <\ 1 $$
-- a contradiction! Thus (in view of Mazur-Ulam Theorem) there does not exist linear isometry $\ J\ $ such that $\ J(w_2) = w_1\ $
End of PROOF
