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 $u,v\in (S,\|\cdot\|_X)$ 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 whether the same holds for general $\|\cdot\|_X$ and $\|\cdot\|_Y.$