How to "scale" the FFT when using it to calculate discrete convolution?
this formula $FFT^{-1} \left( FFT \left( F \right) \odot FFT \left( G \right) \right) \stackrel{!}{=} F \ast G$ is not always true, it's true when you scale at the inverse operation as you mentioned in your first case, in second case you have to multiply the right side by $\sqrt{N}$ to make it right.

Generally when you apply the Fourier transform over the convolution you see any scaling would be used for the first signal and no scaling remain for the second signal so if you multiply and divide by the scaling and use the scaling in the denominator for standard Fourier transform, the scaling remain in the nominator.