Given a circle action on a closed, oriented smooth manifold $M^{2n}$ with isolated fixed points. My question is, does there always exist a point $p\in M$ such that the isotropy subgroup of $p$ is trivial? In other words, does there always exist a free orbit of this circle action? Moreover, if the answer is yes, can we extend this free orbit to a tubular neighborhood $S^1\times D^{2n-1}$ such that each $S^1$ in it is a free orbit?