Can you lift a diffeomorphism $f$ on $S^2$ to a diffeomorphism $F$ on $S^3$ preserving the Hopf fibration structure $\pi:S^3\mapsto S^2$, i.e. such that $\pi\circ F=f\circ \pi$.
I think that if we take a rotation on $S^2$ around the $z$-axis by angle $\theta$ then this can indeed be lifted to a diffeomorphism in $F:S^3\mapsto S^3$ given by $$F(z_1,z_2)=(e^{i\theta}z_1,z_2)$$
But I don't know if this holds for general diffeomorphisms on $S^2$, or at least diffeomorphisms isotopic to the identity, this last case would be enough for me.
So if anyone knows an answer to this I would be deeply grateful.
The Hopf fibration is a fibre bundle with fibre $S^1$. If you have a diffeomorphism $f:M\to N$ and a fibre bundle $\pi_N:P\to N$, you can obtain the pullback bundle $\pi_M:f^*P\to M$. In particular, the pullback bundle is diffeomorphic to the original $P$, because $f$ was a diffeomorphism, so $F:f^*P\to P$, and you have $\pi_N\circ F=f\circ\pi_M$. This is also what is suggested in the comments