If two spheres are isometric, does there exist a bijective isometry $T:S\to S$ with $\|Tu-\alpha Tv\|_Y \leq \|u-\alpha v\|_X$ for all $\alpha>0?$

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.$