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  "body": "\n## Proof\n\nThe key is the formula for conjugating a cycle: conjugation by $\\pi$ applies $\\pi$ to each entry.\n\n1. Let $\\sigma, \\tau \\in S_n$, the [[symmetric group|0RJMNEN3BJ]] on $X = \\{1, \\dots, n\\}$. By the definition of [[conjugate elements|0D1VWQGZZ7]], they are conjugate when $\\tau = \\pi \\sigma \\pi^{-1}$ for some $\\pi \\in S_n$.\n2. Let $c = (a_1\\ \\dots\\ a_k)$ be a [[cycle|0ZKPH2BD1T]] with $k \\ge 1$, and let $\\pi \\in S_n$. We claim $\\pi c \\pi^{-1} = (\\pi(a_1)\\ \\dots\\ \\pi(a_k))$. The points $\\pi(a_p)$ are distinct, since $\\pi$ is [[injective|0PVG1CRAJN]].\n3. Let $y = \\pi(a_p)$. Then $\\pi c \\pi^{-1}(y) = \\pi(c(a_p))$, which is $\\pi(a_{p+1})$ for $p \u003c k$ and $\\pi(a_1)$ for $p = k$.\n4. Let $y$ not be of the form $\\pi(a_p)$. Then $\\pi^{-1}(y)$ is not one of the $a_p$, so $c$ fixes it, and $\\pi c \\pi^{-1}(y) = y$. With step 3, this proves the claim of step 2.\n5. Let $\\rho \\in S_n$, with [[disjoint cycle decomposition|0YWDAZPRPQ]] $\\rho = c_1 \\cdots c_r$. Let $M_i$ be the set of moved points of $c_i$.\n6. For $z \\in M_i$, $c_i(z) \\in M_i$, because $c_i(c_i(z)) = c_i(z)$ would force $c_i(z) = z$ by injectivity. The other factors fix $M_i$ pointwise. So $\\rho(z) = c_i(z) \\ne z$.\n7. A point outside all $M_i$ is fixed by each factor, hence by $\\rho$. So the fixed points of $\\rho$ are exactly the points outside all $M_i$.\n8. Add a $1$-cycle $(a)$ for each fixed point $a$; it is the identity, so the product is still $\\rho$. By step 7, the point sets of all these cycles partition $X$.\n9. By the definition of the [[cycle type|0BCHJ30REG]], the cycle type of $\\rho$ is the multiset of the lengths of the cycles in step 8. We call such a product a full cycle form of $\\rho$.\n10. Assume $\\tau = \\pi \\sigma \\pi^{-1}$. Let $\\sigma = c_1 \\cdots c_s$ be a full cycle form, where $c_i$ has point set $P_i$.\n11. Inserting $\\pi^{-1} \\pi$ between the factors gives $\\tau = (\\pi c_1 \\pi^{-1})(\\pi c_2 \\pi^{-1}) \\cdots (\\pi c_s \\pi^{-1})$.\n12. By step 2, $\\pi c_i \\pi^{-1}$ is a cycle with point set $\\pi(P_i)$ and the same length as $c_i$. Since $\\pi$ is a bijection, the sets $\\pi(P_i)$ partition $X$.\n13. The factors in step 11 of length at least $2$ have pairwise disjoint sets of moved points. By uniqueness in the disjoint cycle decomposition, they are the cycles of $\\tau$ of length at least $2$.\n14. By step 7 for $\\tau$, the remaining points are the fixed points of $\\tau$, and they are the sets $\\pi(P_i)$ of size $1$. So step 11 is a full cycle form of $\\tau$.\n15. By steps 12 and 14, $\\tau$ has the same cycle type as $\\sigma$.\n16. Conversely, assume $\\sigma$ and $\\tau$ have the same cycle type. Take full cycle forms $\\sigma = c_1 \\cdots c_s$ and $\\tau = d_1 \\cdots d_s$, numbered so that $c_i$ and $d_i$ have the same length $k_i$.\n17. Write $c_i = (a_{i,1}\\ \\dots\\ a_{i,k_i})$ and $d_i = (b_{i,1}\\ \\dots\\ b_{i,k_i})$. By step 8, each point of $X$ is exactly one $a_{i,j}$, and exactly one $b_{i,j}$.\n18. So $\\pi(a_{i,j}) = b_{i,j}$ defines a map $\\pi \\colon X \\to X$. The rule $b_{i,j} \\mapsto a_{i,j}$ is a two-sided inverse, so $\\pi$ is [[bijective|01VG2A0BHW]] and $\\pi \\in S_n$.\n19. By step 2, $\\pi c_i \\pi^{-1} = d_i$ for each $i$. As in step 11, $\\pi \\sigma \\pi^{-1} = d_1 \\cdots d_s = \\tau$. So $\\sigma$ and $\\tau$ are conjugate.\n\n## Idea\n\nConjugating by $\\pi$ renames every point $a$ as $\\pi(a)$, so it turns each cycle into a cycle of the same length. Hence conjugation keeps the cycle type. Conversely, if two permutations have the same cycle type, write them cycle by cycle in matching order. The renaming that sends one list of entries to the other conjugates the first permutation to the second.\n",
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