Does the association $V \mapsto GL(V)$ define a functor?
No such functor exists in general. Note that if $n\geq m$, then $k^m$ is a retract of $k^n$, so if such a functor existed, then $GL_m(k)$ would have to be a retract of $GL_n(k)$. I suspect this is basically never true for $n>m>1$, though I don't know a proof for general $k$. In particular, when $k=\mathbb{F}_2$, the group $GL_n(\mathbb{F}_2)$ is simple for all $n>2$, so it has no nontrivial retracts, so such a functor definitely cannot exist for $k=\mathbb{F}_2$.
(Edit 3: The first argument I found is long but there's a shorter argument below it.)
Let's complete Eric's nice argument. First I'll say the first thing he said in a little more detail, which is also what Arturo used. In any category, a retract of an object $c$ is an object $d$ together with two morphisms $f : c \to d, g : d \to c$ such that $fg = \text{id}_d$. This means $f$ is a split epimorphism and $g$ is a split monomorphism. Retracts are the "particularly nice subobjects" (which are also simultaneously quotients); e.g. in an additive category they are precisely the direct summands. The significance of retracts for our purposes is that
- retracts are preserved by arbitrary functors, and
- in $\text{Vect}$, $k^n$ is a retract of $k^m$ iff $n \le m$.
So if $V \mapsto GL(V)$ were a functor, this would imply in particular that $GL_n(k)$ is a retract of $GL_m(k)$ whenever $n \le m$. (Incidentally, for groups retracts correspond exactly to semidirect product decompositions.) So, suppose $f : GL_m(k) \to GL_n(k)$ is an epi and $3 \le n \le m$. It's known that when $m \ge 3$ the subgroup $SL_m(k)$ is perfect, so its image in $GL_n(k)$ must also be perfect and hence must land in $SL_n(k)$. So $f$ induces an epi $SL_m(k) \to SL_n(k)$. An epi maps centers to centers, so $f$ induces an epi $PSL_m(k) \to PSL_n(k)$.
It's also known that when $m \ge 3$, $PSL_m(k)$ is simple. So $f$ is an epi between two nontrivial simple groups, which therefore must be an isomorphism. In other words, we've proven:
Claim: if $V \mapsto GL(V)$ is a functor, then for $3 \le n \le m$, the groups $PSL_n(k)$ and $PSL_m(k)$ are isomorphic.
So it suffices to disprove this.
Edit #2: Okay, I think I can actually complete the argument now. The following weaker result suffices:
Proposition: There exist positive integers $3 \le n < m$ such that $PSL_n(k) \not\simeq PSL_m(k)$.
Proof. We'll proceed by finding $3 \le n < m$ and a nonabelian finite simple group $G$ that embeds into $PSL_m(k)$ but not $PSL_n(k)$. By simplicity, if such a group $G$ embeds into $GL_m(k)$ then any such embedding has image in $SL_m(k)$, and by simplicity again it also embeds into $PSL_m(k)$. Hence if we take $m$ to be the smallest dimension of an irreducible representation of $G$ (which is necessarily faithful) then $G$ embeds into $PSL_m(k)$.
On the other hand, $PSL_n(k)$ acts by conjugation on $\mathfrak{sl}_n(k)$ (the vector space of traceless $n \times n$ matrices), which has dimension $n^2 - 1$ over $k$, so if $G$ embeds into $PSL_n(k)$ then it embeds into $GL_{n^2 - 1}(k)$. Hence if we take $n$ to be such that $n^2 - 1$ is less than the smallest dimension of an irreducible representation of $G$ over $k$, then $G$ does not embed into $GL_{n^2-1}(k)$ and so does not embed into $PSL_n(k)$.
Now it suffices to find a nonabelian finite simple group $G$ such that the smallest dimension of an irreducible representation of $G$ over $k$ is at least $9$ (which means we can take $n = 3$ and $m$ the dimension of this irrep). At this point we'll want to say something about the extent to which this problem is independent of the choice of $k$.
Case 1: $\text{char}(k) = 0$. Any embedding $\rho : G \to GL_d(k)$ of a finite group $G$ into $GL_d(k)$ has image lying in $GL_d(R)$ where $R$ is the $\mathbb{Q}$-subalgebra of $k$ generated by the matrix entries of $\rho(g), g \in G$; since $G$ is finite $R$ is finitely generated, and then by the Nullstellensatz $R$ has a maximal ideal $M$ such that $R/M$ is a number field (a finite extension of $\mathbb{Q}$). By simplicity $G$ also embeds into $GL_d(R/M)$. In other words, the smallest dimension of an irreducible representation of $G$ over an arbitrary field $k$ of characteristic zero is bounded from below by the smallest dimension of an irreducible representation of $G$ over a number field, which is in turn bounded from below by (and is in fact equal to) the smallest dimension of an irreducible representation of $G$ over $\mathbb{C}$. So the characteristic zero case reduces to the case $k = \mathbb{C}$.
At this point we can appeal to standard results; for example it's known that for $n \ge 7$ the alternating group $A_n$ has no nontrivial representations over $\mathbb{C}$ of dimension $\le n-2$. I don't know how to prove this off the top of my head but I think it shouldn't be too hard to at least get a bound of $\le \frac{n-2}{2}$ using standard facts about the representation theory of $S_n$, which would suffice for our purposes.
Explicitly, it follows that $A_{10}$ doesn't embed into $GL_8(\mathbb{C})$ and hence doesn't embed into $PSL_3(k)$, and since it does embed into $PSL_9(k)$ it follows that $PSL_3(k) \not\simeq PSL_9(k)$.
Case 2: $\text{char}(k) = p$ is positive. Repeating the same construction as above, any embedding $\rho : G \to GL_d(k)$ has image lying in $GL_d(R)$ where $R$ is now the $\mathbb{F}_p$-subalgebra of $k$ generated by the matrix entries of $\rho(g), g \in G$. The Nullstellensatz gives that $R$ has a maximal ideal $M$ such that $R/M$ is a finite field $\mathbb{F}_q$ for some $q = p^i$, and we get an embedding $G \to GL_d(\mathbb{F}_q)$. So the characteristic $p$ case reduces to the case $k = \overline{\mathbb{F}_p}$.
In this case we can argue as follows. $GL_d(\mathbb{F}_q)$ has a Sylow $p$-subgroup given by the unipotent subgroup $U_d(\mathbb{F}_q)$ of upper triangular matrices with diagonal entries $1$. This group is nilpotent of class $d-1$, hence all of its subgroups have nilpotency class $\le d-1$. So if $G$ has a Sylow $p$-subgroup of nilpotency class $\ge d$ then it can't embed into $GL_d(\mathbb{F}_q)$. Now it suffices to take $G = PSL_{d+1}(\mathbb{F}_p)$, whose Sylow $p$-subgroup is $U_{d+1}(\mathbb{F}_p)$ (since it lies in the kernel of the determinant map and also has trivial intersection with the center of $SL_{d+1}(\mathbb{F}_p)$).
Explicitly, it follows that $PSL_9(\mathbb{F}_p)$ doesn't embed into $GL_8(\overline{\mathbb{F}_p})$ and hence doesn't embed into $PSL_3(k)$, and since it does embed into $PSL_9(k)$ it follows that $PSL_3(k) \not\simeq PSL_9(k)$. $\Box$
(There's probably a cleaner way to do this...)
Edit 3: There is! We'll show that no functor $F : \text{Vect} \to \text{Grp}$ sending $V$ to $GL(V)$ can exist by using functoriality a bit more.
First, writing $0$ for the zero-dimensional vector space, we have $GL(0) = 1$ (the trivial group). It follows that the zero linear transformation $0 : V \to W$ between any two vector spaces, which factors through $0$, is trivial, since the induced map $F(V) \to F(0) \to F(W)$ factors through the trivial group.
Now consider a short exact sequence $0 \to k^n \to k^{n+m} \to k^m \to 0$ (for example, given by the inclusion of the first $n$ coordinates, and the projection to the last $m$ coordinates). Applying $F$ produces a sequence
$$GL_n(k) \xrightarrow{f} GL_{n+m}(k) \xrightarrow{g} GL_m(k)$$
where $f$ is a split monomorphism, hence in particular injective, and $g$ is a split epimorphism, hence in particular surjective, and $gf$ is trivial. For $n, m \ge 3$ the same conditions hold after passing to commutator subgroups, so we get a sequence
$$SL_n(k) \xrightarrow{f} SL_{n+m}(k) \xrightarrow{g} SL_m(k)$$
where, again, $f$ is injective and $g$ is surjective and $gf$ is trivial. $\text{ker}(g)$ must therefore be a normal subgroup $N$ of $SL_{n+m}(k)$ such that both $N$ and the quotient by $N$ are nonabelian. But by the simplicity of $PSL_{n+m}(k)$ (here $n, m \ge 3$ so $n+m \ge 6$), the only proper normal subgroups of $SL_{n+m}(k)$ are subgroups of its center, which are abelian; contradiction. So no such sequence exists.