I was recently thinking about what it means to put structure on a set. It seems to me that, in my area (representation theory), the two main ways of imposing structure on a set $X$ are:


* distinguishing certain permutations of $X$ as structure preserving;


and


* distinguishing certain test functions (here I think of $X \to \mathbb C$, but that's just because it fits my example) are structure respecting.


For example, given a manifold, we can look at the diffeomorphisms among its permutations, or the smooth functions among its test functions; and, given a vector space, we can look at the linear automorphisms among its permutations, or the functionals among its test functions. Under reasonable hypotheses, the test-functions perspective (more 'analysis'-flavoured, maybe) determines the structure; but does the group perspective (more 'geometry' / Erlangen-flavoured)?


A colleague pointed out that, for manifolds, the diffeomorphism group *does* determine the manifold, in a very strong sense: <http://www.ams.org/mathscinet-getitem?mr=693972>. Given that success, I was moved to ask: does the linear automorphism group determine the vector space?


Here's one way of making that informal question precise. Suppose that $V\_1$ and $V\_2$ are vector spaces (over $\mathbb C$, say) with the same underlying set $X$, and that the set of permutations of $X$ that are linear automorphisms for $V\_1$ is the same as the analogous set for $V\_2$. Then are $V\_1$ and $V\_2$ isomorphic?


Another colleague, and [The Masked Avenger](https://meta.mathoverflow.net/questions/1964/is-it-appropriate-to-ask-a-research-level-question-that-is-not-part-of-a-researc#comment9002_1964), both thought of the axiom of choice when I asked this question; but I'm not sure I see it. It's just a curiosity, so I have no particular investment in whether answers assume, negate, or avoid choice.


EDIT: Since I think it may look like I am making some implicit assumptions, I clarify that I do not mean to assume that the vector spaces are finite dimensional, or that the putative isomorphism from $V\_1$ to $V\_2$ must be the identity as a set map of $X$. Thus, for example, Theo Johnson-Freyd's example (<https://mathoverflow.net/a/186494/2383>) of letting $X$ be $\mathbb C$, and equipping it with both its usual and its 'conjugate' $\mathbb C$-vector space structure, is perfectly OK.



Does the linear automorphism group determine the vector space?
The dimension of $V$ is the least non-negative integer $n$ such that there exist $v\_1,\dotsc, v\_n$ in $V$ such that there exists a unique $g\in G:=GL(V)$ that fixes each of $v\_1,\dotsc,v\_n$. So the isomorphism class of $V$ is determined by the group action of $G$ on $V$.


