In which cases can a metric be defined on a vector space such that there exists an isomorphism to R^n?

I've just learned about vector spaces and I'm curious about for which vector spaces can a metric be defined such that one could create an isomorphism to Euclidean space.

I'm also interested if, in order to define any geometry, it requires the definition of a metric alone, or there have to be other properties.

For example, according to my understanding, Euclidean spaces have a "colloquial" geometric interpretation because they are topological spaces with a metric structure.

My initial hunch was that not all vector spaces can be used to construct a geometry, and for a geometric interpretation to exist the space must have some type of metric structure, or some type of property that formalizes distances.

I would like to know if this thinking is correct or wrong and some guidance on what I should learn/read next to delve into this topic. I'm a Math undergraduate learning linear algebra right now after having learned some group theory, but I haven't learned about topology yet. I might not have the proper terminology, I would just like to learn more about this.

One thing I think isn't emphasised enough at early undergraduate level -- maybe it can't usefully be emphasised at that stage? -- is that natural objects like $\mathbb{R}$ are many things at once. What I mean by that is the following: $\mathbb{R}$ is (1) just a set, (2) a topological space, (3) an abelian group, (4) a ring, (5) a vector space over $\mathbb{Q}$, (6) a well-ordered set...

Which should we think of it as? Well, informally, we tend to think of it as all of these things at once, because some of them are useful for algebra, some are useful for analysis, some are useful for geometry, etc. But they're all subtly different things.

(For example, as a set, $\mathbb{R}$ is isomorphic to $\mathbb{R}\setminus \{\pi\}$, but that makes no sense as abelian groups; however, as abelian groups, $\mathbb{R}$ is isomorphic to $\mathbb{C}$, but that's not true as rings; and you can definitely construct rings that are isomorphic to $\mathbb{R}$ as rings, but not as topological spaces...)

Now, part of what you're asking is "when is a vector space $V$ over a field $F$ isomorphic-as-a-vector-space to $\mathbb{R}^n$ over $\mathbb{R}$?". This bit actually isn't too hard: if you go back and unpack the definition of "isomorphism (of vector spaces)", you'll see that $F$ has to be equal to $\mathbb{R}$ (so if $F = \mathbb{Q}$ or $\mathbb{C}(t)$ or $\mathbb{F}_7$, then $V$ is never isomorphic to $\mathbb{R}^n$); and one of the main goals of a first course in linear algebra is to prove that, when $F = \mathbb{R}$, $$V \cong \mathbb{R}^n \Leftrightarrow \dim_\mathbb{R} V = n.$$

Let's suppose that $f: V\to \mathbb{R}^n$ is an isomorphism of vector spaces, and suppose we have some favourite metric on $\mathbb{R}^n$ already - say, $d_{\mathbb{R}^n}(x,y)$ returns the distance between two points $x,y\in\mathbb{R}^n$. Let's now look at the question in your post title: when can we define a metric on $V$ that means that $V$ is isomorphic to $\mathbb{R}^n$ in some metric space kind of way too? Well, this is actually quite easy too: if $u,v\in V$, then $f(u),f(v)\in\mathbb{R}^n$, so we can just define $d_V(u,v) = d_{\mathbb{R}^n}(f(u),f(v))$. Then $f$ is both an isomorphism of vector spaces and of metric spaces.

So far, so good, but I haven't actually said anything about "geometry" or "Euclidean space" yet. What is our favourite metric on $\mathbb{R}^n$? Well, it's this one - essentially, the distance between two points in space is given by Pythagoras's theorem relative to the axes. Once you've got that, you can define straight lines (e.g. the line between $x$ and $y$ is $\{x + t(y-x) : 0\leq t\leq 1\}$), circles (e.g. as the set of all points a fixed radius from a fixed centre), dot products (from the polarisation identity), angles (e.g. using the cosine rule), etc etc, and you can go ahead and do all of Euclidean geometry from there.

I said above that the distance was defined relative to the axes. Axes are given by a choice of basis. If you change the basis, do you get the same notion of distance? Not necessarily: if you change the basis orthonormally, yes, but it's also possible for basis $B_2$ to look non-orthonormal from the perspective of basis $B_1$ and vice-versa (think: coordinate axes on the plane that aren't at right angles to each other), and this will change the notion of distance. Linear algebra has formalisms for dealing with this too.

Finally, I might as well ramble off topic a little and mention -- since you sometimes mention "geometry" without qualifying it with "Euclidean" -- that there are some natural non-Euclidean geometries too. Spherical geometry is really important if you happen to live on a big sphere, and hyperbolic geometry is important in relativity (Escher liked it too). These aren't vector spaces, I suppose, but both of them can be built from $\mathbb{R}$ (just using slightly different notions of space and distance), and they have some strange properties (e.g. the angles of a triangle don't add up to $180^\circ$ any more!). You might be interested in nonarchimedean geometries, such as those built off the $p$-adic numbers, which also have strange properties (e.g. circles have multiple centres!).

If you're interested in this kind of stuff, I very highly recommend Keith Carne's notes on geometry and groups.