$π, σ ∈ S_{11}$ (permutations of $N_{11} ={1, 2, . . . , 11}$) is given by:
$i:$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $11$
$π(i):$ $9$ $11$ $2$ $3$ $10$ $1$ $4$ $5$ $6$ $8$ $7$
$σ(i):$ $11$ $10$ $1$ $3$ $9$ $5$ $4$ $2$ $6$ $8$ $7$
i) Find the number of $τ ∈ S_{11}$ that satiesfies $τ π = στ$ and state such a $τ$.
ii) Find all $τ ∈ S_{11}$ that satisfy $τ π = στ^{-1}$.
So I have in cycle notation: $π = (1,9,6)(2,11,7,4,3)(5,10,8)$ and $σ = (1,11,7,4,3)(2,10,8)(5,9,6)$.
For i) I have: Define $τ ∈ S_{11}$ where:
$1 ↦ 2$, $9 ↦ 10$, $6 ↦ 8$, $5 ↦ 5$, $10 ↦ 9$, $8 ↦ 6$, $2 ↦ 1$, $11 ↦ 11$, $7 ↦ 7$, $4 ↦ 4$, $3 ↦ 3$ i.e $τ = (1,2)(3)(4)(5)(6,8)(7)(9,10)(11)$ with $τ^{-1} = (11)(10,9)(7)(8,6)(5)(4)(3)(2,1)$. Now this is true since $τ π = στ ↦ τπτ^{-1} = σ$, but it is only one solution. How do I find all solutions and how do I know that they are solutions?
For ii) I have nothing, how do I think here?
In order to find the permutation $ \tau$ that conjugates $\pi$ to $\sigma$, write $\pi$ above $\sigma$ and then read the map $\tau$ off top to bottom \begin{eqnarray*} (1,9,6)& (5,10,8) & (2,11,7,4,3) \\ (2,10,8) & (5,9,6) & (1,11,7,4,3) \end{eqnarray*} giving $\tau=(1,2)(6,8)(9,10)$ as you did in the question. How many ways could we have written $\sigma$ so its cycle shape aligns with $\pi$ ? $2 \times 3 \times 3 \times 5 =\color{red}{90}$.