If there is an into isometry from $(\mathbb{R}^m,\|\cdot\|_p)$ to $(\mathbb{R}^n, \|\cdot\|_q)$ where $m\leq n$, then $p=q$?

In the linked paper is written that Lyubich and Vasertein [9] showed that if there exists an isometric embedding of $(\mathbb{K}^n,\|\cdot\|_p)$ to $(\mathbb{K}^m,\|\cdot\|_q)$ where $\Bbb K=\Bbb R$ or $\Bbb K=\Bbb C$ and $1 < p, q <\infty$ then $p =2$, $q$ is an even integer, and $m$ [I guess such minimal $m$. AR.] satisfies the inequality $${n+q/2-1\choose n-1}\le m\le {n+q-1\choose n-1}.$$

For instance, it can be checked that a linear map $T:(\mathbb{R}^2,\|\cdot\|_2)\to (\mathbb{R}^3, \|\cdot\|_4)$ such that $T(0,1)=(a,b,c)$ and $T(1,0)=(b,a,-c)$, where $b=\sqrt{\frac{2\sqrt{2}-\sqrt{6}}{6}}$, $a=(2+\sqrt{3})b$, and $c=\frac{\sqrt[4]{2}}{\sqrt{3}}$, is an isometry, because $(a,b,c)$ is a solution of a system of equations $$\cases{ a^4+b^4+c^4=1\\ a^3b+b^3a-c^4=0\\ 12a^2b^2+6c^4=2}.$$