Is $\mathbb{R}^n$ isomorphic to itself when the metric used is $d^2(x,y) = \|x-y\|^2$ rather than the standard euclidean metric?

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I am wondering if the metric space $(\mathbb{R}^n, d)$, where $d(x,y) = \|x-y\|$ is the standard euclidean metric, is isomorphic to itself under the squared euclidean metric: $d^2(x,y) = \|x-y\|^2$.

That is, for all pairs of points $x,y \in \mathbb{R}^n$, does there exist a bijection, $f : \mathbb{R}^n \to \mathbb{R}^n$ such that $d(x,y) = d^2(f(x),f(y))$ ?

More generally, is it true if we consider metrics derived from arbitrary p-norms, $d^p(x,y) = \|x-y\|^p$, $p\ge 1$ ?

Finally, if the answers to the above questions are "no", then can we at least get a "yes" for some values of $n$ (where $n \ge 1$)?

Or can we get a "yes" by relaxing the requirement that $f$ be a bijection from $\mathbb{R}^n$ to itself, and instead allow $f$ to embed points in some other number of dimensions, "m", i.e. $f : \mathbb{R}^n \to \mathbb{R}^m$, with $d(x,y) = d^p(f(x),f(y))$ ?