Nonfree orbits of a smooth circle action with finite stabilizers have pairwise coprime orders Suppose we are given a smooth circle action on a smooth compact manifold $X$ such that the stabilizer of every point of $X$ is finite (so all orbits are circles). Assume that there are only finitely many nonfree orbits. (According to https://www.intlpress.com/site/pub/files/_fulltext/journals/pamq/2008/0004/0002/PAMQ-2008-0004-0002-a001.pdf (p.6), such an action is called pseudo-free.) Then is it true that the stabilizers of the nonfree orbits have pairwise coprime orders?
I am asking this question because I found a different definition of a pseudo-free circle action (first page of https://arxiv.org/pdf/0904.2975.pdf), which says that a pseudo-free circle action is a smooth circle action that is free except for finitely many nonfree orbits whose isotropy types have pairwise coprime orders.
 A: As indicated in this MO question, there is a circle action on the Klein bottle which is free except for two exceptional orbits where the isotropy groups are both $\mathbb{Z}/2\mathbb{Z}$.  In particular, they are not relatively prime.
If you want a more geometric understanding of this action, here's how I think about it.  Start with the usual $S^1$ action on $S^2$.  This action commutes with the antipodal action, so descends to an $S^1$ action on $\mathbb{R}P^2$.  I'm going to think about $\mathbb{R}P^2$ as the northern hemisphere of $S^2$ with equatorial antipodal points identified.  Thinking of $\mathbb{R}P^2$ in this way, the circle action is free except that it fixes the north pole and has $\mathbb{Z}/2\mathbb{Z}$ isotropy at the equator.
Now, take another copy of $\mathbb{R}P^2$ with this action.  We will carefully form the connect sum in such a way as to retain the circle action.  Specifically, the $S^1$ action on $\mathbb{R}P^2$ preserves everything south of the artic circle.  So, the balls we'll remove from both copies consist of everything north of the artic circle.  Notice that we've now removed the fixed points.
Put the two copies of $\mathbb{R}P^2\setminus{ball}$ next to eacho ther with the boundaries adjacent.  If the circle action rotates the boundary circles in opposite directions, reverse the direction of the $S^1$ action on one copy.  Now we can glue equivariantly, finding an $S^1$ action on $\mathbb{R}P^2\sharp \mathbb{R}P^2$ which is free except at the two "equators", where the isotropy is $\mathbb{Z}/2\mathbb{Z}$.  Finally, simply note that the Klein bottle is diffeomorphic to this connect sum.
