Does every metric space that topologically embeds into $\mathbb R^n$ isometrically embeds into some metrizable space homeomorphic to $\mathbb R^n$?

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Follow up to this question. Motivated by wanting to extend a metric space into one that intuitively "extends indefinitely and has no gaps". And also motivated by the famous result that every metrizable space embeds into some $\mathbb R^\kappa$:

Question: Given a metric space $M$ that topologically embeds into some $\mathbb R^n$ for finite $n$, is there always a metric space $N$ that is topologically $\mathbb R^n$ for some finite $n$ (perhaps the same $n$?) such that $M$ isometrically embeds into $N$?

An answer for $n = \aleph_0$ would also be interesting, but I'm mostly interested in the finite $n$ case.