Colored chirality of knots

For a given knot diagram $K$ with a (Fox) $n$-coloring ${\mathcal C}$, its mirror $K^*$ has the $n$-coloring ${\mathcal C}^*$ (called the mirror coloring) that is the mirror image of ${\mathcal C}$. Specifically, let ${\mathcal A}$ be the set of (over-)arcs of a diagram $K$, then the mirror image $K^*$ has the set of arcs ${\mathcal A}^*$ that is in a natural bijection with ${\mathcal A}$, such that each arc $a \in {\mathcal A}$ has its mirror $a^* \in {\mathcal A}^*$, and vice-versa (for any arc $\alpha \in {\mathcal A}^*$, there is a unique arc $a \in {\mathcal A}$ such that $\alpha=a^*$). Then ${\mathcal C}^* : {\mathcal A}^* \rightarrow \mathbb{Z}_n$ is defined by ${\mathcal C}^* (a^*)={\mathcal C} (a)$.

Figure: Is figure-eight amphicheiral with $5$-colorings?

In Fig.  a $5$-colored figure-eight knot diagram and its mirror with its mirror coloring are depicted. It is, then, natural to ask if an $n$-colored knot diagram (a pair $(K, {\mathcal C})$ of a diagram $K$ and a coloring ${\mathcal C}$) is equivalent to its mirror with its mirror coloring $(K^*, {\mathcal C}^*)$. If this is the case, we call the colored diagram $(K, {\mathcal C})$ amphicheiral (or achiral) with (respect to) the $n$-coloring ${\mathcal C}$. Otherwise we say a diagram is chiral with (respect to) the $n$-coloring.

The answer to the above question is NO, and it is seen from the contributions of the $3$-cocycle invariant with the dihedral quandle $R_5$ are distinct between the left and right of Fig. . Many other achiral knots have colorings with which they become chiral.