The trefoil exterior is Seifert-fibered

The action preserves $\mathrm{\mid }z{\mathrm{\mid }}^2+\mathrm{\mid }w{\mathrm{\mid }}^2$, so it acts smoothly on $S^3$. The orbit of $(2^{-1\mathrm{/}2},2^{-1\mathrm{/}2})$ is exactly $K$. A group element stabilizes this point only if $3\theta$ and $2\theta$ are both multiples of $2\pi$; coprimality of $2$ and $3$ makes the stabilizer trivial. The regions $\mathrm{\mid }z\mathrm{\mid }\le \mathrm{\mid }w\mathrm{\mid }$ and $\mathrm{\mid }w\mathrm{\mid }\le \mathrm{\mid }z\mathrm{\mid }$ are solid tori, parametrized respectively by $(z,w)=(z,\sqrt{1-\mathrm{\mid }z{\mathrm{\mid }}^2}\eta )$ and $(z,w)=(\sqrt{1-\mathrm{\mid }w{\mathrm{\mid }}^2}\eta ,w)$ with the disk coordinate bounded by $2^{-1\mathrm{/}2}$ and $\eta \in S^1$. The $w$ coordinate is the core direction of the first solid torus and winds twice along $K$; the $z$ coordinate is the core direction of the second and winds three times. This proves the claimed $(3,2)$ torus-knot type. On the axis $z=0$ the stabilizer has order $2$, and on the axis $w=0$ it has order $3$.

On $zw\mathrm{\ne }0$, write $\alpha =\arg z$ and $\beta =\arg w$ modulo $2\pi$. The integer coordinate change $\theta =\alpha -\beta$, $\varphi =2\alpha -3\beta$ has determinant $-1$ and is invertible on the phase torus. The circle action changes $\theta$ by its parameter and fixes $\varphi$, while $r=\mathrm{\mid }z\mathrm{\mid }\in (0,1)$ is fixed; thus the quotient of this open set is the cylinder with coordinates $(r,\varphi )$, and the inverse image of any disk $D$ around $q(K)$ in it is explicitly $S_{\theta }^1\times D$. At $r=0$ and $r=1$ the phase circle collapses to one point at each end, so the full orbit space is a sphere; the two axis orbits have stabilizer orders $2$ and $3$. The complement of $D$ is a disk with those two cone points. Its orbits give the fibers of $E_K$, and [F2] identifies the axis fibers as exceptional of orders $2$ and $3$.

Choose a basepoint $x$ on a regular fiber, let $h$ be that fiber loop, and let $\gamma$ be any loop in $E_K$ based at $x$. The map $F:[0,1{]}^2\to E_K$ defined by $F(s,t)=g_{2\pi s}(\gamma (t))$ is well-defined because the tube is invariant. Its boundary represents $h\gamma h^{-1}{\gamma }^{-1}$: the two vertical sides are $\gamma$ and ${\gamma }^{-1}$ since $g_0=g_{2\pi }=\mathrm{id}$, and the horizontal sides are the fiber loop and its reverse. The square contracts its boundary, so $h$ commutes with $[\gamma ]$. Since $\gamma$ was arbitrary, the regular fiber class is central in ${\pi }_1(E_K)$.

By [F3], a Seifert-fibered exterior does not satisfy the required non-Seifert-fibered hypothesis. Step 2.1 establishes that $E_K$ is Seifert-fibered, so this example refutes the statement that every knot exterior is non-Seifert-fibered and also fails that hypothesis of the hyperbolization contract.