Power-of-two periodicity conjecture for pattern-avoidance counts

Let $f_n^k(\sigma)$ denote the number of permutations of length $n$ associated with a pattern $\sigma\in S_3$ and fixed by the $k$th iterate of the fundamental bijection. Power-of-two periodicity conjecture. For $i,j\geq2$, $n\geq1$, and $\sigma\in S_3$,

$$f_n^{2^i}(\sigma)=f_n^{2^j}(\sigma).$$

This conjecture proposes that all iterates whose exponents are powers of two with exponent at least two give the same count for every length and every pattern in $S_3$; the supplied text gives no proof or resolution.