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).

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 $PT = S$.

The answer is no in general. E.g., let $X=\mathbb R$ with the standard norm and $Y=\mathbb R^2$ with the norm given by the formula $\|y\|=|y_1|\vee|y_2-y_1|$. Let $S$ and $T$ be defined by the formulas $S1:=(1,0)$ and $T1:=(0,1)$. Then $\|Tx\|=\|Sx\|$ for all $x\in\mathbb R$.

If $P$ is an isometry of $Y$ such that $PS=T$, then $P(1,0)=(0,1)$ and $P(0,1)=(a,b)$, where $\|(a,b)\|=1$. Further, then would have $|y_1|\vee|y_2-y_1|[=\|y\|=\|Py\|]=|y_2a|\vee|y_1+y_2(b-a)|$ for all real $y_1,y_2$.

But this is impossible for any real $a,b$. Indeed, letting $y_1=y$ and $y_2=1$, we would have $|y|\vee|y-1|=|a|\vee|y+b-a|$ for all real $y$. Letting here $y\to\infty$, we would get $y=y+b-a$, so that $b=a$ and hence $|y|\vee|y-1|=|a|\vee|y|$ for all real $y$. Letting here $y\to-\infty$, we would get $1-y=-y$, a contradiction.