Orientation-reversing involution on $S^3$, commuting with a circle action?

Question

Consider $S^1$ as a subset of $\mathbb{R}^2\cong\mathbb{C}$, and $S^3$ as a subset of $\mathbb{R}^4\cong\mathbb{C}^2$, and define an action of $S^1$ on $S^3$ by $z\cdot(w_1,w_2):=(zw_1,zw_2)$. Does there exist a smooth orientation-reversing involution $F:S^3\to S^3$ which commutes with the action, i.e. which satisfies $F(z\cdot(w_1,w_2))=z\cdot F(w_1,w_2)$ for all $z\in S^1$ and $(w_1,w_2)\in S^3$?

Answer 1

Through the magic of Smith Theory, we know that the fixed point set of an orientation reversing involution on $S^3$ is either $S^0$ or $S^2.$ If this involution commutes with a free $S^1$ action, then the fixed point set must admit a free $S^1$ action, but neither $S^0$ nor $S^2$ does. So there ain't no such animal.