Does $S^4$ admit a smooth $S^3$-action?
If there is such an action, I think it would have to be non-free: if it were free, then the quotient manifold theorem gives that $S^4\to Q:=S^4/S^3$ is an submersion. By the long exact sequence of homotopy groups, $Q$ would be a compact simply connected 1-manifold, i.e. the interval $[0,1]$. But $[0,1]$ is contractible, so an $S^3$ bundle over $[0,1]$ would have to be trivial. But $S^4\neq S^3\times [0,1]$, so the action cannot be free.
Now $S^4$ is "almost" $S^3\times [0,1]$ in the sense that $S^4$ is the quotient of $S^3\times [0,1]$ with the sets $S^3\times \{0\}$ and $S^3\times \{1\}$ collapsed to a point.
Pick a nontrivial smooth self-action of $S^3$ on itself (maybe conjugation?). Write coordinates on $S^4$ locally as $(h,t)$ where $h\in S^3$ and $t\in [0,1]$. Define $f:S^3\times S^4\to S^4$ by $$(g,(h,t))\mapsto (ghg^{-1},t).$$
This action is trivial for $t=0,1$, so it is not free. I'm not sure if it is smooth. Does this example work?
More generally, suppose $S^3\to M\to N$ is a smooth $S^3$-bundle with $M$ compact. Does $M$ then necessarily admit a nontrivial smooth $S^3$-action?