I have an easier solution not involving solution of differential equations, the type Griffiths himself might prefer.

The force on the charge is only due to the field of the induced charge on the cavity inner surface. If you go to page 101, 2 pages behind the question itself, he states the the field of the induced charge and the field of the main charge alone cancel inside the conductor. For a spherical cavity this is obvious that the leftover charge on the outer surface has zero field inside the conductor, but it holds for all shapes.

Now the potential in the cavity due to the induced charge must be same everywhere in the cavity for there to be no force. So $\mathbf{E}$ must be zero for this case in the cavity, with no charge in the cavity (main charge has no force on itself due to it's own field) a gaussian surface returns a null integral of field over surface. Now only two ways exist, either $\mathbf{E}$ is zero over the entire gaussian surface, which is possible only if we can assume symmetry of E, but that is only when cavity is spherical all other unsymmetrical configurations will have positive and negative finite values over part of the surface to make this integral zero. So it is proven that E is not zero if the cavity is not spherical, and that proves the potential is non uniform. That proves force is zero.