If I have an abelian group $(V,+)$ and a field $K$, then I can make a vector space if I define a product $·\space\colon K \times V \longrightarrow V$ which is associative, distributive w.r.t. both the sum of $V$ and the sum of $K$, and satisfies $1·\vec{v}=\vec{v}$ for any $\vec{v}\in V$.
If I'm given an abelian group and a field, how many vector spaces can I make with them?, which is to say, how many products make $V$ a vector space? I have no clue of how should I approach this.
I know, though, that if we know of a valid product $\lambda·\vec{v}$, then $\lambda * \vec{v} = \sigma(\lambda)·\vec{v}$ is also a valid product for any field automorphism $\sigma$ on $K$... so maybe I should count all products "up to field isomorphism of $K$".
Answer
For prime fields, a finite dimensional vector space is uniquely determined by its abelian group structure. This includes finite fields of prime order and the rational numbers.
However, every finite dimensional vector space over the real numbers is isomorphic as an abelian group to any other, and is also isomorphic to a real vector space of countably infinite dimension. Also, a vector space of dimension $mn$ over a field with $p$ elements is isomorphic as an abelian group to a vector space of dimension $n$ over a field with $p^m$ elements. In particular, any vector space over a field $F$ with a subfield $E$ is also a vector space over $E$.
As you observed, field automorphisms also mess with uniqueness, but this is actually well studied. A map between vector spaces that is linear except possibly twisted by a field automorphism is called sesquilinear, and you can search for info on that on the internet.
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