An isometry from metric space $X=\mathbb{R}^{d}$ to metric space $Y=\mathbb{R}^{k}$ with usual norm for both spaces is the following:
$$ \Phi: \mathbb{R}^{d} \rightarrow \mathbb{R}^{k} $$ where $\Phi(x)=Px$ and $P \in \mathbb{R}^{k \times d}$ such that $\|x\|_2^2=\|Px\|_2^2$. The set of all $P$'s can be written as
$$ \mathcal{P}=\{P \in \mathbb{R}^{k \times d} \mid \|x\|_2^2=\|Px\|_2^2 \,\,\,\, \forall x \in \mathbb{R}^{d}\} $$
My questions:
Is $\mathcal{P}$ the set of all $P \in \mathbb{R}^{k \times d}$ where each column is a Euclidean basis $e_i=[0,0,\cdots,0,\overbrace{1}^{i},0,\cdots,0]^{T}$?
If so, why this is true and we have only zero and 1 in the isometry?