For the record, let me write down the Mehler kernel for the square root of the Fourier transform, $$K(x,y)=(2\pi)^{-1/2}\sqrt{1-i}\exp\left[{\tfrac{1}{2} i \left(x^2+y^2-2 \sqrt{2} x y\right)}\right].$$ As a check, I calculate $$\int_{-\infty}^\infty K(x,z)K(z,y)\,dz=(2\pi)^{-1/2}e^{-ixy},$$ so indeed, application of the transform $K$ twice is equivalent to the Fourier transform.

For the record, to answer Q2, let me write down the Mehler kernel for the square root of the Fourier transform, $$K(x,y)=(2\pi)^{-1/2}\sqrt{1-i}\exp\left[{\tfrac{1}{2} i \left(x^2+y^2-2 \sqrt{2} x y\right)}\right].$$ It satisfies $$\int_{-\infty}^\infty K(x,z)K(z,y)\,dz=(2\pi)^{-1/2}e^{-ixy},$$ so application of the transform $K$ twice is equivalent to the Fourier transform, defined as $${\cal F}(x)=(2\pi)^{-1/2}\int_{-\infty}^\infty e^{-ixy}f(y)\,dy.$$