Let me give a very partial answer, in line with my comment. If $y^2=x^3+1$ is supersingular, i.e. $p\equiv5\pmod6$, then the cube roots of unity are not in the prime field, so that the automorphism $(x,y)\mapsto(\omega x,y)$ does not commute with Frobenius $(x,y)\mapsto(x^p,y^p)$.

Similarly, if $y^2=x^3-x$ is supersingular, i.e. $p\equiv3\pmod4$, then the fourth roots of unity are not in the prime field, and the automorphism $(x,y)\mapsto(-x,iy)$ does not commute with Frobenius.

I’m sure that in each case, the four endomorphisms you see form a $\Bbb Q$-basis for $\Bbb Q\otimes_{\Bbb Z}\mathrm{End}$, and it looks to me as if they ought to form a $\Bbb Z$-basis for the endomorphism ring itself.

I don’t know how to handle primes that are $\equiv1\pmod{12}$.