If two spheres are isometric, does there exist a bijective isometry $T:Sto S$ with $|Tu-alpha Tv|_Y leq...
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Let $$(S,|cdot|) = {(x,y)in mathbb{R}^2: |(x,y)| =1},$$ that is, $S$ is the collection of all norm one vectors in $mathbb{R}^2$ with respect to the norm $|cdot|.$ Question : Let $|cdot|_X$ and $|cdot|_Y$ be two norms on $mathbb{R}^2$ be such that $(S,|cdot|_X)$ and $(S,|cdot|_Y)$ are isometric. Does there exist a bijective isometry $T:(S,|cdot|_X)to (S,|cdot|_Y)$ such that $$|Tu-alpha Tv|_Y leq |u-alpha v|_X$$ for all $u,vin (S,|cdot|_X)$ and all $alpha>0?$ Note that the norms $|cdot|_X$ and $|cdot|_Y$ may be distinct. I tried $|(x,y)|_X = |x|+|y|,$ $|(x,y)|_Y = max{|x|,|y|}$ and $$T(x,y) = begin{pmatrix} 1 & 1 \ -1 & 1 end{pmatrix}.$$ Note that $T$ is a rotation matrix. Clearly $T$ is a bijective isometry and satisfies the inequality. However, I do not know wheth...