A space isometric to $\ell_\infty^2$ Consider a norm on $\mathbb C^2$ as $\|(z_1,z_2)\|:=\max\{|z_1|,|z_2|,\frac{1}{\sqrt{2}}|z_1+iz_2|\}.$
Question. Is $(\mathbb C^2,\|.\|)$ linearly isometric to $(\mathbb C^2,\|.\|_{\infty})$ where $\|(z_1,z_2)\|_\infty:=\max\{|z_1|,|z_2|\}?$
 A: There is no such map $f$. Let's try to map from the second space (with the funny norm, which I'll denote simply by $\|\cdot\|$) back to $(\mathbb C^2, \|\cdot \|_{\infty})$. Let $u=f(e_1)$, $v=f(e_2)$, so $\|u\|_{\infty}=\|v\|_{\infty}=1$. Since $|1\pm i|^2=2$, so $\|(1,\pm 1)\|=1$, we also have $\|u\pm v\|_{\infty}=1$. However, if $|z|=1$ and $w\not= 0$, then $|z\pm w|>1$ for one choice of sign. So if (say) $|u_1|=1$, then $v_1=0$. It follows that $u=e^{i\alpha}e_1$, $v=e^{i\beta}e_2$, or the other way around.
But then $\|f((1,-i)/\sqrt{2})\|_{\infty}=1/\sqrt{2}$ even though $\|(1,-i)/\sqrt{2}\|=1$.
A: comment
I think they are not isometric, having different structure for the set of extreme points.
The set of extreme points for the unit ball of $\|\cdot\|_\infty$ is a torus: $$T = \{(z_1,z_2) : |z_1| = |z_2| = 1\}.$$
The set of extreme points of the unit ball of $\|\cdot\|$ perhaps consist of the union of three tori: $$T_1 = \{(z_1,z_2) : |z_1|=|z_2| = 1\},\\T_2 = \textstyle\{(z_1,z_2) : |z_1|=\frac{1}{\sqrt{2}}|z_1+iz_2| = 1\},\\T_3 = \{(z_1,z_2) : |z_2|=\textstyle\frac{1}{\sqrt{2}}|z_1+iz_2| = 1\}.$$
Here is an example of a point in $T_2$ but not $T_1$:
$$
\left(1,\;
\frac{\sqrt{-r^4+6r^2-1}}{2} + i\,\frac{r^2-1}{2}\right),\quad
\sqrt{2}-1 < r < 1 .
$$

Now all that remains is to prove this...
