We first give a bijection between tables $C \in \mathcal{F}$ and partitions of a fixed regular hexagon of side length $n$ into parallelograms given by two unit equilateral triangles glued together. Call such a partition a well-partitioned hexagon. For a well-partitioned hexagon, align one of its edges parallel to the $y$ axis so that the hexagon lies to the right side of this edge and divide it into $n$ unit edges.

Consider the first unit edge on the top, which is an edge of a unit parallelogram. Connect the midpoint $M_0$ of this edge to the midpoint $M_1$ of the opposite edge of this parallelogram. We write 1 in the (1,1) cell of an empty $n \times 2n$ table $C$ if $M_0 M_1$ has positive slope, and 0 otherwise. Similarly, we take $M_2$ to be the next midpoint and write 1 or 0 in the (1,2) cell if $M_1 M_2$ has positive slope or not, respectively. Iterate this step $2n$ times to fill the first row of $C$ with 0's and 1's. We do the same thing for the second unit edge on the left edge of $H$ to fill the second row of $C$, and so on. The result is a $n \times 2n$ table whose cells are filled with 0 or 1. An example of this correspondence is below.

$$C = \begin{bmatrix} 1 & 1 & 0 & 0 & 1 & 0 \\ 1 & 0 & 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 & 1 & 1 \end{bmatrix} \longleftrightarrow \text{Hexagon}$$

Since the height of the $k$-th unit edge on the left edge of $H$ is the same as the $k$-th one on the right edge of $H$, the number of 0's and 1's in the $k$th row is the same, namely $n$. If $a_{k,i} < a_{k+1,i}$ for some $k, i$, let $i_0$ be the minimum of such $i$'s. Then among the $1, 2, \dots, a_{k,i_0} - 1$st columns, the number of 0's in the $k$th and $k+1$st row are the same. Hence, the $a_{k,i_0} - 1$st edge of the $k$th row is adjacent to the $a_{k,i_0} - 1$st edge of the $k$-th and $k+1$-th row. But then the next parallelograms of the $k$-th and $k+1$-th row overlap, a contradiction. Hence, we have that $C \in \mathcal{F}$. Similarly, one can check that for any $C \in \mathcal{F}$, one can find a corresponding well-partitioned hexagon $H$.

We are now ready to address both parts of the problem.

a. A rotation of a well-partitioned hexagon is still well-partitioned, so $f(C)$ corresponds to a well-partitioned hexagon, hence lies in $\mathcal{F}$.

b. The well-partitioned hexagon associated to $f^6(C)$ is simply a rotation of the hexagon associated to $C$ by 360 degrees, hence is the same as the hexagon associated to $C$. Therefore, by our bijection, we find that $C = f^6(C)$.