A bijection on permutations

Question

I am looking for a bijection between permutations in $\mathfrak S_n$ with a certain weight and a second set, which arises by interpreting the expression $$ \frac{1}{2}\left(1 + \exp(q \log\left(\frac{1+x}{1-x}\right)\right) $$ as the exponential generating function for permutations with only odd cycles, where each cycle is coloured blue or red, up to interchanging the colours. Each cycle carries the weight $q$. For example, there are $4!=24$ cycles of length 5, $\binom{5}{2}\cdot 2\cdot 2^2 = 80$ coloured permutations with a 3-cycle and two singletons, and $2^4 = 16$ coloured permutations with 5 singletons.

On the other hand, the weight of a permutation in $\mathfrak S_n$ is obtained by computing https://www.findstat.org/StatisticsDatabase/St000389oMp00093oMp00127oMp00066oMp00090 : write down the cycle decomposition of the permutation, cycles ordered by minimal elements, and such that the minimal element of each cycle comes first. Then erase the parenthesis and interpret the result as a permutation $\sigma$ in one-line notation. Finally, interpret this permutation as a Dyck path - drawn as a subdiagonal path from $(0,0)$ to $(n,n)$ - with peaks at $\{(i, \sigma_i-1) | \sigma_i\text{ is a right-to-left minimum of }\sigma\}$. For example, the permutation $2,4,3,1$ has cycle decomposition $(1,2,4)(3)$, thus $\sigma=1,2,4,3$, so the Dyck path has peaks at $(4,3-1)$, $(2,2-1)$, and $(1, 1-1)$. Each ascent of odd length of this Dyck path carries the weight $q$.

This question arises by comparing this answer with this, and the following comments.

Answer 1

In what follows I describe a bijection between your colored odd cycle only permutations and all permutations. Under this bijection, the number of cycles of the initial odd cycle only permutation becomes the number of cycles of the output permutation plus some additional factor (which could be described but seems a bit complicated). I am not sure how this weight relates to your proposed weight— see my comments below the answer.

Start with a permutation with only odd cycles, and interpret the red/blue coloring as a binary string of length equal to the number of cycles minus one.

Write the permutation in “standard cycle form” with the biggest element of each cycle first and with biggest elements increasing from left to right. E.g., (3)(746)(95182)

Then process the binary word by starting at the leftmost cycle and whenever we see a one move the last element of that cycle to be the last element of the cycle to its right (whereas do nothing when we see a zero). Proceed to the next cycle when processing the next letter, and so on.

E.g., with (3)(746)(95182) and the word 11 we process the first 1 and get (7463)(95182), then we process the second 1 and end with (746)(951823).

The inverse is obtained by starting with any permutation and moving the last elements of the cycles (in standard form) from right to left so as to make sure the cycles are all of odd length.

The basic idea behind this bijection appears in Section 6.2 of Bóna’s “Walk Through Combinatorics” textbook (3rd ed.) and is explained in the answers to the previous MO question: Permutations with all cycles odd length and permutations with all cycles even length.