Trivial colouring and the fundamental group of a knot

Question

Suppose that a classical knot diagram $K$ is coloured by a finite quandle $X$ such that only one element in the quandle is used to colour the knot $K$. We say that the colouring is trivial in this case. Is the follwoing statement true: If the knot admit only trivial colouring by any finite quandle, then can we conclude that the fundmental group of the knot is isomorphic to $\mathbb{Z}$?

Answer 1

For a quick review of part of Joyce's 1980 paper: Given a quandle $Q$ and a group $G$, there is a quandle $\operatorname{Conj} G$ whose operation is conjugation ($g\triangleright h:=h^{-1}gh$), and there is a group $\operatorname{Adconj} Q$ generated by the elements of $Q$ with the relations $y^{-1}xy=x\triangleright y$ for all $x,y\in Q$. Quandle maps $Q\to \operatorname{Conj}H$ are in one-to-one correspondence with group homomorphisms $\operatorname{Adconj Q}\to H$.

If every quandle map $Q\to R$ with $R$ a finite quandle is trivial, then it follows that for every finite group $H$, every group homomorphism $\operatorname{Adconj}Q\to H$ has abelian image.

For knots, the fundamental quandle's Adconj is the fundamental group of the knot complement. Thus, we may assume the finite quotients of $\pi_1(S^3-K)$ are all abelian. Since the abelianization is $H_1(S^3-K)\cong\mathbb{Z}$, all the finite quotients are in fact cyclic, and furthermore the finite quotients consist of all the finite cyclic groups. The only group whose finite quotients are all the finite cyclic groups is $\mathbb{Z}$ itself due to its profinite rigidity. (So $K$ must be the unknot.)