Generally, I am interested in understanding the free actions of finite cyclic groups on $S^3$ which leave invariant an oriented torus knot $T(p,q)$. For a specific example, consider the knot $T(3,4)$ and the group $\mathbb{Z}/5\mathbb{Z}$. My understanding is that there is a unique (up to equivariant homoemorphism of pairs $(S^3,K)$) order 5 symmetry of each torus knot, see for example the top of page 2 here, since there is a unique order 5 subgroup of $S^1$. However, here are two free order 5 symmetries of $T(3,4)$, where $H$ is a right-handed half twist, $F^2$ is two right-handed full twists, $I$ is the trivial tangle, $-F$ is a left-handed full twist, and $T(3,4)$ is the closure of the shown braid:

$1/5$th of the way to the right, and then doing $1/5$th of a left-handed full twist." />,

and

$1/5$th of the way to the right, and then doing $2/5$ths of a right-handed full twist." />.

It seems to me that in the first example the quotient of $S^3$ is the lens space $L(5,-1)$ and in the second example the quotient is the lens space $L(5,2)$. Since these 3-manifolds are not homeomorphic, there then could not be an equivariant homemomorphism of pairs $(S^3,T(3,4))$ between these two symmetries. This seems to contradict the uniqueness of the order 5 symmetry on $T(3,4)$.

Is there indeed a unique order 5 symmetry of $T(3,4)$? If so, up to what equivalence of $\mathbb{Z}/5\mathbb{Z}$ symmetries? Can I see directly that these two symmetries are the same under that equivalence?

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