I have an easier solution not involving solution of differential equationsLet us assume that the conductor and cavity are asymmetric random shapes with $q$ placed at any general location in the cavity. This charge q, will induce a charge -q on the type Griffiths himself might prefercavity surface. The leftover charge after -q is drawn is +q and that distributes itself over the conductor exterior surface.
The forceForce on the charge, $F=q/vec(E)$ is only due to thebe made zero. So Electric field at the location of the induced charge onmust be zero, assuming that it's a general point, we need the field to be zero at all points of the cavity inner surface. If you goNow, what is producing this field?, When calculating forces, the field due to the othercharges is only important, so we can assume q to be removed entirely and calculate field due to -q and +q. On page 101, 2 ehich is two pages behind the question itselfproblem, heGriffiths clearly states thethat the third field of the induced charge anddue to +q is separately zero than the fieldsum of the main charge alone cancel insideq and -q in the conductor. For a spherical cavitySo we neglect +q in this calculation, this is obvious thatunderstopd by the leftover chargefact that conductors have infinite permittivity leading to symmetrical (zero) field on the outera gaussian surface has zero field insidein the conductor, but it holds for all shapes.
Now the potential in and as the cavityintegral due to it is zero as the induced charge must be same everywheresurface does not enclose it, and as such distance in the cavityconductors for there to befield is of no force. So $\mathbf{E}$ must be zero for this case in the cavity,consequence as it does not decrease with no charge indistance, a symmetrical field will give zero integral that is the cavity (main charge has no force on itselffield due to it's own+q is always zero. This is a logical and not mathematical proof.
The field) is now only due to -q. We have already assumed the cavity as asymmetrical. Taking a gaussiangauss integral ( surface returns a null integral of electric field over surfaceand dot product with perpendicular) is equal to the charge enclosed (zero) divided by permittivity. Now only two ways existfor the integral to be zero there are three possibilities, eitherthat the $\mathbf{E}$$/vec(E)$ is zero overat all points of the entire gaussian surface, which is possible only if we can assume symmetry of E, but that (i) or it is only when cavityperpendicular to the surface (ii) or it is spherical all other unsymmetrical configurations will haveequally positive andas it is negative finite values over part ofon the surface (iii). As we are free to make thisassume any shape of the surface for gauss integral irrespective of cavity (assuming no limitation due ease of calculation) we can simply take a spherical shape that would make the curl of E non zero fo case (ii), which rules out (ii). So itField can only be zero at all points for all gaussian surfaces for a symmetrical surface, specifically a sphere. Thus (iii) is the only possibilty and as proven that E is not zero ifby the cavity is not sphericalsecond uniquemess theorem, if there is a field and charge distribution related to each other, you can be sure that proves the potentialit is non uniform. That proves forcea totally one to one relation for all space considered (not for localized areas, as is zerothe case for method of images).
