Let us consider $S^1$ as a Lie group, and suppose we are given a smooth $S^1$-action on a closed manifold.
According to this paper (https://www.jstor.org/stable/60608), the action is said to be pseudo-free if it is not free and if every orbit is one-dimensional and if the isotropy group is the identity except on isolated exceptional orbits where the isotropy group is the finite cyclic group $\Bbb Z_k$, $k > 1$.
But, according to this paper (https://arxiv.org/pdf/0904.2975.pdf), the action is said to be pseudo-free if it is free except for finitely many non-free orbits whose isotropy types $\Bbb Z_{m_1},\dots , \Bbb Z_{m_n}$ have pairwise relatively prime orders.
Are these two definitions equivalent? (I'm expecting this, because these two papers are related to each other: the first paper is in the reference list of the second one.) If the action satisfies the second condition then it clearly satisfies the first condition too. But, is the coverse also true? If not, I can't see what is the motivation in the second definition that the isotropy groups should have pairwise relatively prime orders.