Solution:

The answer is (C). We show that, in moving the first constrained vertex of the triangle from one end to the other of a single side of the hexagon, the unconstrained vertex travels back and forth along a segment of length $\frac{2 \sqrt{3}}{3}-1$.

Indeed, let $A B C D E F$ be the hexagon, let $O$ be its center; let $G$ and $H$ be on segments $A B$ and $B C$, respectively, such that $G H=A B$, and let $R$ be the point on the same side of $O$ with respect to the line $G H$ such that $G H R$ is an equilateral triangle. We want to show that $R$ lies on the line $O B$.

We note that the quadrilateral $G B H R$ is cyclic, since the sum of the angles $\widehat{G B H}=120^\circ$ and $\widehat{G R H}=60^\circ$ is a straight angle. It follows that $\widehat{G B R}=\widehat{G H R}=60^\circ$, hence $\widehat{G B R}=\widehat{G B O}$ as desired. We also compute the distance of $R$ from $O$ in the case where $G H$ is orthogonal to $B O$ (that is, $H$ is the reflection of $G$ with respect to that line). In this case, letting $K$ be the point of intersection between $B O$ and $G H$, the triangles $K G B$ and $K H B$ are congruent and right-angled, each being half of an equilateral triangle, hence $B K=\frac{1}{\sqrt{3}}$, $G K=\frac{\sqrt{3}}{6}$. Since $R K=\frac{\sqrt{3}}{2}$ and $O B=1$, we have $R O=R K-(O B-B K)=2 \frac{\sqrt{3}}{3}-1$.

It follows that, as a constrained vertex of the right triangle moves from $A$ to $B$ and the second constrained vertex from $B$ to $C$, the unconstrained vertex travels back and forth along a segment $O X$ of length $2 \frac{\sqrt{3}}{3}-1$. In total, the length traveled by the unconstrained vertex is $12 \cdot O X=8 \sqrt{3}-12$.