Okay, me previous question Can a vector space be reconstructed form its norm? was solved by a nice (but in hindsight embarassingly simple) counter-example. Perhaps a better question emerges from that asnwer as well as from the discussions in the comments there:
Let $(V,+,\cdot,\|\cdot\|)$ be a normed $\Bbb R$ vector space. Assume we are only given the underlying set $V$, the zero vector $0$, and the induced metric $d\colon V\times V\to\Bbb R$, $(v,w)\mapsto\|v-w\|$. Can we reconstruct $+$, $\cdot$, and $\|\cdot\|$ from this?
The last part is of course trivial because $\|v\|=d(v,0)$. On the other hand, if we were not given $0$ as a base point, we'd be doomed because $d$ is translation invariant. (Or we might instead be satisfied with reconstructing the affine structure)
At least in case of the Euclidean norm on a finite-dimensional space, the answer is Yes:
First, we can determine $\dim V$ as one less than the maximal number of vertices in a regular simplex. If we pick such a maximal simplex with side length $1$, say, and with one vertex $0$, then we can "triangulate" all other points of $V$ by their distances to the vertices, and this allows us to reconstruct $+$ and $\cdot$.
Remains the question, what the situation looks like in all other cases. Actually, this single question automatically splits into two parts:
Q1: What is the situation with other norms on finite-dimensional spaces? "Triangulation" becomes a non-trivial task and even finding a "nice" simplex may prove difficult.
Q2: What is the situation with infinite dimensional spaces? Those can be even weirder …
When pressed to pick one, I'd be more interested in answers to Q1 …
EDIT: To clarify, added $\Bbb R$ as scalar field.
Best Answer
I'll assume the scalar field is $\mathbb R$. The Mazur-Ulam theorem asserts that any surjective isometry of normed linear spaces is affine. So (up to choice of origin) the metric does uniquely determine the vector space structure.