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?



a question about the isotropy subgroup of circle action on manifolds with isolated fixed point
The answer is `no', but for a stupid reason: you can have an action with an 
 ineffective kernel, meaning that (normal closed) subgroup of $G=S^1$
consisting of those $g$ which act trivially: for all $x$ in $M$, $gx = x$. For example,
take a free action of $S^1$ on $M$. Define a new action $g \* x = g^p x$
(I'm thinking of the circle multiplicatively). The new action's ineffective kernel consists of the pth roots of unity. 


You can get rid of the ineffective kernel by taking $G$ and
dividing by the ineffective kernel. In this circle case, the resulting action of this `new' 
$S^1$ will be free a.e.: that is it will have principal orbit type the identity (check out, for example, Hsiang's book,`Cohomology Theory of Topological Transformation Groups', p. 10-12, esp. p.12) which means, combined with the `slice theorem' (p.10) or
the`Gleason theorem' (p. 9) the answer to your question becomes `yes'. 


