Does the "coordinate-to-basis-vector" functor have an adjoint? Consider global coordinate systems on $\mathbb R^n$. Such a coordinate system is defined by a list of $n$ smooth functions
$$x^i: \mathbb R^n \to \mathbb R, \quad i = 1,... n$$
such that their gradients at every point make up a basis (that is, their contour surfaces are fully transverse at every point).
Also consider global frame systems on $\mathbb R^n$. Such a frame system is defined by a list of $n$ smooth vector fields
$$v^i: \mathbb R^n \to \mathbb R^n, \quad i = 1,... n$$
such that they make up a basis at every point.
The global coordinate systems make up a category $\mathcal C$, with coordinate transforms being the arrows. Similarly, the global frame systems make up a category $\mathcal F$, with smooth pointwise linear transforms being the arrows.
Then, we have a contravariant functor $V: \mathcal C \to \mathcal F$, which sends $\{x^1, ... x^n\}$ to $\{\partial_{x^1}, ... \partial_{x^n}\}$, and sends coordinate transform $y^i = f^i(x^1, ... x^n)$ to linear transform $\partial_{x^i} = (\partial_{x^i}f^j)\partial_{y^j}$.
There are strictly more frame systems than coordinate systems, because given the frame system corresponding to a coordinate system, we can integrate it to recover the coordinate system. However, there are many frame systems that cannot be integrated.

 For example, in $\mathbb R^2$, consider this frame system:
 $$\{(1 + a^2 a^2) \hat e_1, \hat e_2 \}\quad\text{at point } (a^1, a^2)$$
 There is no way to consistently assign a coordinate to the point $(1, 1)$, since if you follow the path $(0, 0) \to (0, 1) \to (1, 1)$, you would move by $(1, 1)$. But if you follow the path $(0, 0) \to (1, 0) \to (1, 1)$, you would move by $(0.5, 1)$.

So $V$ is an injective functor, like free functors. Free functors often have right adjoints called forgetful functors.
So the question is: does $V$ have a right adjoint $W: \mathcal F \to \mathcal C$, a forgetful functor that "forgets" a frame system into a coordinate system?
(Of course, $V$ is contravariant, so it would have to actually be a functor $W: \mathcal F^{op} \to \mathcal C$, but you get the point.)
 A: Yes, and the right adjoint is whatever functor you want :D However, this is due to the particular nature of involved categories more than the free-forgetting yoga.
Firstly we want to show that for all $z, w \in \mathcal{F}$ it holds
$$ Hom_{\mathcal{F}}(z, w) = \{*\}$$
i. e. it has cardinality one. Indeed, recall that a global frame $z$ is given by vectors $z_1(x), \ldots, z_n(x) $ that spans $\mathbb{R}^n$ smoothly varying with $x$. We denote by $Z(x) $ the matrix obtained by putting the columns $z_1(x), \ldots, z_n(x) $ one after another. A morphism from z to w is a smoothly varying $L(x) $ such that for every $i$
$$L(x) z_i(x) = w_i(x) $$
Which can be rewritten as
$$ L(x) Z(x) = W(x) $$
Since $Z(x) $ is invertible, this implies
$$ L(x) = W(x) Z(x) ^{-1}$$
Also, notice that the inverse is a smooth function, since it can be written as the adjoint matrix (polynomial) multiplied by 1 over the determinant (smooth where different by zero). This gives us a unique morphism.
Now we notice that this extends to $\mathcal{C}$. As you noticed, the two functors $V, I$ exhibits $Hom_{\mathcal{C}}(z, w) $ as a non empty subset of $Hom_{\mathcal{F}}(Vz, Vw) $; thus it has cardinality one, too.
Let's pass to the adjointness part. We will show that any functor from $\mathcal{F}$ to $\mathcal{C}$ is adjoint to $V$. Take any functor $F$, and define the natural transformations as the only morphisms $1 \to FV, VF \to 1$. Commutation rules are always satisfied, because when you have to verify if two morphism are equal, you fall into an hom set of cardinality one, so they must be the same.
My suggestion: to get something more meaningful, you should relax a bit the conditions on morphisms. In particular, it would be nice if naturality was more demanding to get, and if global frames morphisms were not all isomorphisms. For example, if you allow "singular frames" to exist, you could have non invertible transformations too.
Hope I didn't mislead what you meant!
