Let $n>3$ be a positive integer.We denote the symmetric group of $n$ elements by $S_n$ and the identity mapping by $id$. For every $f\in S_n$, $f(1,2,\ldots,n)=(a_1,a_2,\ldots,a_n)$, denote $a_n$ by $m(f)$.

For any positive integer $1\leq k\leq n-2$, define $f_k\in S_n$ as follow:

$f_k:(1,\ldots,k,k+1,k+2,\ldots,n-1,n)\to (1,\ldots,k,n,n-1,\ldots,k+2,k+1)$.

Then I conjecture that if $i_0,i_1,\ldots,i_l\in \{1,2,\ldots,n-2\}$ satisfy:

$(1)l>1$;

$(2)i_0=1$;

$(3)$For any $1\leq u\leq v\leq l$ such that $\{u,u+1,...,v\}$ is a proper subset of $\{0,1,...,l\}$,$f_{i_v}\circ ...\circ f_{i_{u+1}}\circ f_{i_u}\neq id$;

$(4)f_{i_l}\circ \ldots\circ f_{i_1} \circ f_{i_0}=id$,

we must have $\{m(f_{i_0}),m(f_{i_1}\circ f_{i_0}),\ldots,m(f_{i_l}\circ \ldots\circ f_{i_1}\circ f_{i_0})\}=\{2,3,\ldots,n\}$.

Is it true? If not, please give a counterexample.