Question regarding why capacitance is a purely geometrical quantity To show that the capacitance (of some capacitor), defined as $C \equiv Q/V$, depends solely on the geometry of the capacitor, we use that the electric field is proportional to the charge density, $\rho$, and that $\rho$ at some point is proportional to $Q$. But as Griffiths points out,

Wait a minute! How do we know that doubling $Q$ (and also $-Q$) simply doubles $\rho$? Maybe the charge moves around into a completely different configuration, quadrupling $\rho$ in some places and halving it in others, just so the total charge on each conductor is doubled.

He then explains that

The fact is that this concern is unwarranted—doubling $Q$ does double $\rho$ everywhere; it doesn't shift the charge around. The proof of this will come in Chapter 3; for now you'll just have to trust me.

What is this proof Griffiths is referring to?
Edit: Chapter 3 largely deals with existence and uniqueness theorems concerning Poisson's equation. It seems obvious to me that the proof is closely related to those theorems.
 A: In my copy (3rd edition), on p. 118, §3.1.6, is the following theorem:

Second uniqueness theorem: In a volume $\mathcal V$ surrounded by conductors and containing a specified charge density $\rho$, the electric field is uniquely determined if the total charge on each conductor is given (Fig. 3.6). (The region as a whole can be bounded by another conductor, or else unbounded.)


For a capacitor, the charge density $\rho$ in the sense of this theorem is zero, and the region as a whole is unbounded. The theorem makes it very clear that there is one and only one solution corresponding to each set of charges on the conductors. Thus, if you have one field configuration $\mathbf E(\mathbf r)$ corresponding to charges $\pm Q$ on the two plates of a capacitor, then it is immediately obvious what is the (unique) field configuration that corresponds to charges $\pm 2 Q$ on those two same plates: it is simply $2\mathbf E(\mathbf r)$, i.e. twice your original solution.
Moreover, we know that the charge on the conductors is concentrated at the boundaries, and that at those boundaries it has the surface charge density
$$
\sigma = \epsilon_0 \hat{\mathbf n}\cdot \mathbf E,
$$
i.e. the normal component at the boundary. Since we know the electric field for the up-scaled configuration, this also gives us the charge configuration at all points on the surfaces of the conductors.
