Counter-example concerning isometry. Can $(\Bbb R^2 , \|\cdot\|_1)$ and $(\Bbb R^2, \|\cdot\|_2)$ be isometric?

I think that these two spaces can't be isometric. But I can't able to show that. Would anybody please help me in this regard?
Thanks in advance.
(Apologies, I cannot comment). This question is a duplicate, and already has answers here.
Assume that $U:(\mathbb{R}^d,\|\cdot\|_1)\to(\mathbb{R}^d,\|\cdot\|_2)$ is an isometry. Then
\begin{align*}
\|Ux+Uy\|_2^2+\|Ux-Uy\|_2^2&amp;=2\|Ux\|_2^2+2\|Uy\|_2^2
\end{align*}
by the parallelogram law. On the other hand, as $U$ is an isometry,\begin{align*}
\|x+y\|_1^2+\|x-y\|_1^2&amp;=2\|x\|_1^2+2\|y\|_1^2.
\end{align*}
Thus $\|\cdot\|_1$ also satisfies the parallelogram law, thus comes from a scalar product, a contradiction. The conclusion is the same if we replace $\|\cdot\|_1$ by any $\|\cdot\|_p$ with $p\neq2$.