Let $X,Y$ be finite-dimensional real normed spaces. Consider the set of linear operators $L(X,Y)$ between the two spaces.


Then we define the set of equivalence classes 
$$G(X,Y):=\left\{[T]; T,S \in L(X,Y) \text{ are equiv., if there is an isometry P on Y with }PT=S \right\}.$$


I ask: Does $$\sup\_{x \in X} \left\lvert \left\lVert Tx \right\rVert-\left\lVert Sx \right\rVert \right\rvert =0$$ imply that $[T]=[S]$?



The geometry of $\mathbb{R}^n$
The answer is no, in general.


In order to construct a counterexample, let $X = Y = \mathbb{R}^n$ for any $n \ge 2$ and endow this space with the $p$-norm for your favourite $p \in [1,\infty] \setminus \{2\}$. The point about this choice of the norm is that the (linear) isometries on $\mathbb{R}^n$ with respect to this norm are exactly the signed permutation matrices, i.e. permutation matrices for which some of the one's have been replaced with $-1$ (see [this post](https://mathoverflow.net/questions/288052/what-are-the-matrices-preserving-the-ell1-norm/288084#288084)).


Now, fix an arbitrary linear functional $\varphi: \mathbb{R}^n \to \mathbb{R}$ of norm $1$. Let $e\_1 \in \mathbb{R}^n$ be the first canonical unit vector and let $f \in \mathbb{R}^n$ be a vector of norm $1$ such that neither $f$ nor $-f$ is a canonical unit vector. We define $Tx = \langle \varphi, x\rangle e\_1$ and $Sx = \langle \varphi, x\rangle f$ for each $x \in \mathbb{R}^n$. 


Then we have $\|Tx\| = \|Sx\| = |\langle \varphi, x \rangle|$ for all $x \in \mathbb{R}^n$. However, if we had $PT = S$ for an isometry $P$, this would imply $\langle \varphi,x\rangle P e\_1 = \langle \varphi, x \rangle f$ for all $x \in \mathbb{R^n}$. By plugging in any $x \in \mathbb{R}^n$ which is not in the kernel of $\varphi$, we conclude that $Pe\_1 = f$. But his cannot be true since $P$ is a signed permutation matrix.


**EDIT** If think it's worthwhile to add the following general result which answers the question in a more complete way:


**Theorem 1.** The answer to the question is "yes" if and only if $Y$ is (isometrically) a Hilbert space.


For the proof, we need the following result:


**Theorem 2.** A finite dimensional Banach space $Y$ is isometrically a Hilbert space if and only if the group of all linear isometries on $Y$ acts transitively on the unit sphere of $Y$ (i.e. for all vectors $y\_1,y\_2 \in Y$ of norm $1$ we can find a linear isometry $P$ on $Y$ such that $Py\_1 = y\_2$).


It is interesting to note that it is open whether the same result remains true for separable Banach spaces; see for instance the article [Ferenczi and Rosendal: On isometry groups and maximal symmetry (2013)] for details. The finite dimensional result in Theorem 2 is attributed to Mazur in [op. cit., page 1773].


**Proof of Theorem 1.** "$\Rightarrow$" Assume that $Y$ is not a Hilbert space. By Theorem 2 we can find two vectors $y\_1,y\_2 \in X$ such that $Py\_1 \not= y\_2$ for any isometry $P$ on $Y$. Now, one can mimic the construction from the first part of this answer by replacing the vectors $e\_1$ and $f$ with $y\_1$ and $y\_2$, respectively.


"$\Leftarrow$" Let $Y$ be a Hilbert space. Since $\|Tx\| = \|Sx\|$ for all $x,y \in X$, the operators $T$ and $S$ have the same kernel $K$. Choose a vector subspace $V \subseteq X$ such that $X = K \oplus V$. Then the restrictions $T|\_V: V \to Y$ and $S|\_V: V \to Y$ are injective and the mapping $J := S|\_V \;
 (T|\_V)^{-1}: \operatorname{rg}T \to \operatorname{rg}S$ is a bijective isometry from the range of $T$ to the range of $S$.


Since $\operatorname{rg}T$ and $\operatorname{rg}S$ have the same dimension, so do their orthogonal complements in $Y$. Thus, there exists an isometry $R: (\operatorname{rg}T)^\perp \to (\operatorname{rg}S)^\perp$. The operator $P := J \oplus R$ is an isometry on $Y$ which fulfils $PT = S$.


