This question is inspired by a game I played.
Define an action $P_k$ on a finite sequence $\{a_n\}_{n=1}^m (m>k)$: choose $k$ consecutive terms (from the sequence), move them to the middle of two terms (or append them at the beginning or end) of the remaining sequence to form a new sequence $\{a_n'\}_{n=1}^m$.
For example, if we apply $P_3$ to the sequence $1,3,7,5,6$, the result could be \begin{align*} 1,3,\color{red}{7,5,6}&\xrightarrow{P_3}1,\color{red}{7,5,6},3\\ 1,3,\color{red}{7,5,6}&\xrightarrow{P_3}\color{red}{7,5,6},1,3\\ 1,\color{blue}{3,7,5},6&\xrightarrow{P_3}1,6,\color{blue}{3,7,5}\\ 1,\color{blue}{3,7,5},6&\xrightarrow{P_3}\color{blue}{3,7,5},1,6\\ &\quad\vdots \end{align*}
Question
1. Prove: any sequence formed by permutations of $1,2,3,4,5,6$ could be "reduced" to $1,2,3,4,5,6$ after applying $P_3$ for finite times.
2. Find all $k$ such that any sequence formed by permutations of $1,2,3,\dots,2k$ could be "reduced" to $1,2,3,\dots,2k$ after applying $P_k$ for finite times.
For Qn 1, let $\mathcal B=\{(1\ 2),(2\ 3),(3\ 4),(4\ 5),(5\ 6)\}$, where all elements in $\mathcal B$ are cycle notation. For example, $(1\ 2)$ means $\color{red}{a_4,a_6},a_5,a_2,a_1,a_3\to\color{red}{a_6,a_4},a_5,a_2,a_1,a_3$.
Apply $P_3$ several times, we could get $(1\ 2)$: \begin{align*} \begin{array}{ccccccccc} 1\color{red}{234}56&\to &\color{red}{234}156&&&&&&\\ &=&234\color{blue}{156}&\to&2\color{blue}{156}34&&&&\\ &&&=&21\color{orange}{563}4&\to&214\color{orange}{563}&&\\ &&&&&=&21\color{green}{456}3&\to&213\color{green}{456} \end{array} \end{align*} By symmetry, we could get $(5\ 6)$. We could also get $(2\ 3)$ and $(3\ 4)$ with some effort. Since all $6!=720$ permutaions could be generated by the composition of elements in $\mathcal B$ (e.g. $(1\ 3)=(2\ 3)(1\ 2)(2\ 3)$), we finished the proof.
For Qn 2, I tried to use the same approach, but failed. I think we might choose a different $\mathcal B$ such as $\mathcal B':=\{(1\ 2),(1\ 3),(1\ 4),\dots, (1\ 2k)\}$.
Are there any elegant ways to solve these questions?