TITLE: Does the linear automorphism group determine the vector space?
QUESTION [16 upvotes]: 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, 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.

REPLY [8 votes]: Vladimir Dotsenko is right:

$GL(V)\cong GL(W)$ iff $V\cong W$. 

This is true for any complex (and not only complex) vector spaces. 
If at least one of the dimensions is finite, then this follows from Dotsenko's argument. If $\dim V$ and $\dim W$ are infinite, this is valid for vector spaces over arbitrary division rings by
Tolstykh's theorem [Theorem 8.3] (the paper contains a complete proof of the 
theorem and a remark that it "easily follows from general isomorphism theorems proved by O’Meara in [15, Theorem 5.10, Theorem 6.7]").
[15] O. O’Meara, A general isomorphism theory for linear groups, J. Algebra, 44 (1977) 
93-142.