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Here is a different viewpoint: the question is equivalent to asking for a permutation $\sigma$ whose disjoint cycle structure contains exactly $m$ cycles, and with $|C_{S_{n}}(\sigma)|$ as small as possible subject to that.

Note that if $\sigma$ has $a_{j}$ cycles of length $j$ for each $j$ in that decomposition, then $|C_{S_{n}}(\sigma)| = \prod_{j} j^{a_{j}} a_{j}!,$ (which is just the denominator in the expression given in the question for the size of the conjugacy class). Note that this is clearly at least as large as $ \prod_{ \{j : a_{j} >0\}} j$, and note on the other hand that equality holds if all cycle sizes are distinct.

From now on I consider the case where there is a partition of $n$ into $m$ parts of distinct sizes (which is equivalent to requiring that $n \geq \frac{m(m+1)}{2}).$ I want to show first that there is a unique (up to conjugacy) choice of $\sigma$ which minimizes $|C_{S_{n}}(\sigma)|$ subject to the condition that $\sigma$ has $m$ cycles of pairwise distinct sizes.

Given such a permutation $\sigma,$ choose $j = j(\sigma)$ maximal subject to the fact that the $r$-th shortest cycle of $\sigma$ has size $r$ whenever $r \leq j$ - the possibility that $j =0$ is allowed, and occurs precisely when the shortest cycle of $\sigma$ has size greater than one. Note that the $j+1$-st shortest cycle of $\sigma$ (if there is one) has size at least $j+2$ by the maximality of $j$.

If $\sigma$ has more than $j+1$ disjoint cycles, then we may produce a permutation $\tau$ with $m$ disjoint cycles of pairwise distinct lengths and with $|C_{S_{n}}(\tau)| < |C_{S_{n}}(\sigma)|.$ We simply shorten the $j+1$-st shortest cycle of $\sigma$ by one, and lengthen the longest cycle of $\sigma$ by one. The resulting permutation $\tau$ still has all its $m$ disjoint cycles of pairwise distinct lengths, but we see easily that if $a$ is the length of the $j+1$-st shortest cycle of $\sigma$ and $b > a $ is the length of the longest cycle of $\sigma,$ then $ab|C_{S_{n}}(\tau)|= (b+1)(a-1)|C_{S_{n}}(\sigma)|$, so the claim follows as $(b+1)(a-1) < ab.$

Hence we can "improve" our choice of $\sigma$ if $\sigma$ has more than one cycle of length greater than $j+1,$ so the unique minimizing choice of $\sigma$ has its $r$-th cycle of length $r$ for $r \leq j$ and only one cycle of length greater than $j+1$. This forces $j = m-1$ or $j =m.$

Hence if $n \geq \frac{m(m+1)}{2},$ the unique choice of $\sigma$ with $m$ disjoint cycles of pairwise distinct lengths which minimizes $|C_{S_{n}}(\sigma)|$ has its $r$-th cycle of length $r$ for $1 \leq r \leq m-1$ and one cycle of length $n - \frac{m(m-1)}{2}.$ Note that the size of the corresponding conjugacy class is $\frac{n!}{(n-\frac{m(m-1)}{2})(m-1)!}$.

While this does not directly answer the question, it does perhaps give some insight and provides large conjugacy classes of elements with $m$ disjoint cycles.

Here is a different viewpoint: the question is equivalent to asking for a permutation $\sigma$ whose disjoint cycle structure contains exactly $m$ cycles, and with $|C_{S_{n}}(\sigma)|$ as small as possible subject to that.

Note that if $\sigma$ has $a_{j}$ cycles of length $j$ for each $j$ in that decomposition, then $|C_{S_{n}}(\sigma)| = \prod_{j} j^{a_{j}} a_{j}!,$ (which is just the denominator in the expression given in the question for the size of the conjugacy class). Note that this is clearly at least as large as $ \prod_{ \{j : a_{j} >0\}} j$, and note on the other hand that equality holds if all cycle sizes are distinct.

From now on I consider the case where there is a partition of $n$ into $m$ parts of distinct sizes (which is equivalent to requiring that $n \geq \frac{m(m+1)}{2}).$ I want to show first that there is a unique (up to conjugacy) choice of $\sigma$ which minimizes $|C_{S_{n}}(\sigma)|$ subject to the condition that $\sigma$ has $m$ cycles of pairwise distinct sizes.

Given such a permutation $\sigma,$ choose $j = j(\sigma)$ maximal subject to the fact that the $r$-th shortest cycle of $\sigma$ has size $r$ whenever $r \leq j$ - the possibility that $j =0$ is allowed, and occurs precisely when the shortest cycle of $\sigma$ has size greater than one. Note that the $j+1$-st shortest cycle of $\sigma$ (if there is one) has size at least $j+2$ by the maximality of $j$.

If $\sigma$ has more than $j+1$ disjoint cycles, then we may produce a permutation $\tau$ with $m$ disjoint cycles of pairwise distinct lengths and with $|C_{S_{n}}(\tau)| < |C_{S_{n}}(\sigma)|.$ We simply shorten the $j+1$-st shortest cycle of $\sigma$ by one, and lengthen the longest cycle of $\sigma$ by one. The resulting permutation $\tau$ still has all its $m$ disjoint cycles of pairwise distinct lengths, but we see easily that if $a$ is the length of the $j+1$-st shortest cycle of $\sigma$ and $b > a $ is the length of the longest cycle of $\sigma,$ then $ab|C_{S_{n}}(\tau)|= (b+1)(a-1)|C_{S_{n}}(\sigma)|$, so the claim follows as $(b+1)(a-1) < ab.$

Hence we can "improve" our choice of $\sigma$ if $\sigma$ has more than one cycle of length greater than $j+1,$ so the unique minimizing choice of $\sigma$ has its $r$-th cycle of length $r$ for $r \leq j$ and only one cycle of length greater than $j+1$. This forces $j = m-1$ or $j =m.$

Hence if $n \geq \frac{m(m+1)}{2},$ the unique choice of $\sigma$ with $m$ disjoint cycles of pairwise distinct lengths which minimizes $|C_{S_{n}}(\sigma)|$ has its $r$-th cycle of length $r$ for $1 \leq r \leq m-1$ and one cycle of length $n - \frac{m(m-1)}{2}.$ Note that the size of the corresponding conjugacy class is $\frac{n!}{(n-\frac{m(m-1)}{2})(m-1)!}$.

While this does not directly answer the question, it does perhaps give some insight and provides large conjugacy classes of elements with $m$ disjoint cycles.

Later edit: As indicated in the last sentence, and suggested also by comments of Gerhard Paseman, this is not at all the end of the story if we allow repeated cycle lengths.

For example, if $\sigma$ has cycle type $12 \ldots (m-2)(m-1)d$ where d > m, we can shorten (m-2) cycles of $\sigma$ by one each, and lengthen the longest cycle by m-2 to obtain $\tau$ of cycle type $112 \ldots (m-2)(d+m-2)$.

Then $(m-1)d|C_{S_{n}}(\tau)| = 2(d+m-2)|C_{S_{n}}(\sigma)|$ which yields $|C_{S_{n}}(\tau)| < |C_{S_{n}}(\sigma)|$ as long as $m>3.$ Then $\tau$ is the product of $m$ disjoint cycles, but $\tau$ has a larger conjugacy class than $\sigma$. This suggests other continuations, and seems to point towards some of the empirically found maxima found by Aaron in his answer, at least when $n$ is large enough compared to $m.$

Here is a different viewpoint: the question is equivalent to asking for a permutation $\sigma$ whose disjoint cycle structure contains exactly $m$ cycles, and with $|C_{S_{n}}(\sigma)|$ as small as possible subject to that.

Note that if $\sigma$ has $a_{j}$ cycles of length $j$ for each $j$ in that decomposition, then $|C_{S_{n}}(\sigma)| = \prod_{j} j^{a_{j}} a_{j}!,$ (which is just the denominator in the expression given in the question for the size of the conjugacy class). Note that this is clearly at least as large as $ \prod_{ \{j : a_{j} >0\}} j$, and note on the other hand that equality holds if all cycle sizes are distinct.

From now on I consider the case where there is a partition of $n$ into $m$ parts of distinct sizes (which is equivalent to requiring that $n \geq \frac{m(m+1)}{2}).$ I want to show first that there is a unique (up to conjugacy) choice of $\sigma$ which minimizes $|C_{S_{n}}(\sigma)|$ subject to the condition that $\sigma$ has $m$ cycles of pairwise distinct sizes.

Given such a permutation $\sigma,$ choose $j = j(\sigma)$ maximal subject to the fact that the $r$-th shortest cycle of $\sigma$ has size $r$ whenever $r \leq j$ - the possibility that $j =0$ is allowed, and occurs precisely when the shortest cycle of $\sigma$ has size greater than one. Note that the $j+1$-st shortest cycle of $\sigma$ (if there is one) has size at least $j+2$ by the maximality of $j$.

If $\sigma$ has more than $j+1$ disjoint cycles, then we may produce a permutation $\tau$ with $m$ disjoint cycles of pairwise distinct lengths and with $|C_{S_{n}}(\tau)| < |C_{S_{n}}(\sigma)|.$ We simply shorten the $j+1$-st shortest cycle of $\sigma$ by one, and lengthen the longest cycle of $\sigma$ by one. The resulting permutation $\tau$ still has all its $m$ disjoint cycles of pairwise distinct lengths, but we see easily that if $a$ is the length of the $j+1$-st shortest cycle of $\sigma$ and $b > a $ is the length of the longest cycle of $\sigma,$ then $ab|C_{S_{n}}(\tau)|= (b+1)(a-1)|C_{S_{n}}(\sigma)|$, so the claim follows as $(b+1)(a-1) < ab.$

Hence we can "improve" our choice of $\sigma$ if $\sigma$ has more than one cycle of length greater than $j+1,$ so the unique minimizing choice of $\sigma$ has its $r$-th cycle of length $r$ for $r \leq j$ and only one cycle of length greater than $j+1$. This forces $j = m-1$ or $j =m.$

Hence if $n \geq \frac{m(m+1)}{2},$ the unique choice of $\sigma$ with $m$ disjoint cycles of pairwise distinct lengths which minimizes $|C_{S_{n}}(\sigma)|$ has its $r$-th cycle of length $r$ for $1 \leq r \leq m-1$ and one cycle of length $n - \frac{m(m-1)}{2}.$ Note that the size of the corresponding conjugacy class is $\frac{n!}{(n-\frac{m(m-1}{2})(m-1)!}$.

While this does not directly answer the question, it does perhaps give some insight and provides large conjugacy classes of elements with $m$ disjoint cycles.

Here is a different viewpoint: the question is equivalent to asking for a permutation $\sigma$ whose disjoint cycle structure contains exactly $m$ cycles, and with $|C_{S_{n}}(\sigma)|$ as small as possible subject to that.

Note that if $\sigma$ has $a_{j}$ cycles of length $j$ for each $j$ in that decomposition, then $|C_{S_{n}}(\sigma)| = \prod_{j} j^{a_{j}} a_{j}!,$ (which is just the denominator in the expression given in the question for the size of the conjugacy class). Note that this is clearly at least as large as $ \prod_{ \{j : a_{j} >0\}} j$, and note on the other hand that equality holds if all cycle sizes are distinct.

From now on I consider the case where there is a partition of $n$ into $m$ parts of distinct sizes (which is equivalent to requiring that $n \geq \frac{m(m+1)}{2}).$ I want to show first that there is a unique (up to conjugacy) choice of $\sigma$ which minimizes $|C_{S_{n}}(\sigma)|$ subject to the condition that $\sigma$ has $m$ cycles of pairwise distinct sizes.

Given such a permutation $\sigma,$ choose $j = j(\sigma)$ maximal subject to the fact that the $r$-th shortest cycle of $\sigma$ has size $r$ whenever $r \leq j$ - the possibility that $j =0$ is allowed, and occurs precisely when the shortest cycle of $\sigma$ has size greater than one. Note that the $j+1$-st shortest cycle of $\sigma$ (if there is one) has size at least $j+2$ by the maximality of $j$.

If $\sigma$ has more than $j+1$ disjoint cycles, then we may produce a permutation $\tau$ with $m$ disjoint cycles of pairwise distinct lengths and with $|C_{S_{n}}(\tau)| < |C_{S_{n}}(\sigma)|.$ We simply shorten the $j+1$-st shortest cycle of $\sigma$ by one, and lengthen the longest cycle of $\sigma$ by one. The resulting permutation $\tau$ still has all its $m$ disjoint cycles of pairwise distinct lengths, but we see easily that if $a$ is the length of the $j+1$-st shortest cycle of $\sigma$ and $b > a $ is the length of the longest cycle of $\sigma,$ then $ab|C_{S_{n}}(\tau)|= (b+1)(a-1)|C_{S_{n}}(\sigma)|$, so the claim follows as $(b+1)(a-1) < ab.$

Hence we can "improve" our choice of $\sigma$ if $\sigma$ has more than one cycle of length greater than $j+1,$ so the unique minimizing choice of $\sigma$ has its $r$-th cycle of length $r$ for $r \leq j$ and only one cycle of length greater than $j+1$. This forces $j = m-1$ or $j =m.$

Hence if $n \geq \frac{m(m+1)}{2},$ the unique choice of $\sigma$ with $m$ disjoint cycles of pairwise distinct lengths which minimizes $|C_{S_{n}}(\sigma)|$ has its $r$-th cycle of length $r$ for $1 \leq r \leq m-1$ and one cycle of length $n - \frac{m(m-1)}{2}.$ Note that the size of the corresponding conjugacy class is $\frac{n!}{(n-\frac{m(m-1)}{2})(m-1)!}$.

While this does not directly answer the question, it does perhaps give some insight and provides large conjugacy classes of elements with $m$ disjoint cycles.

Here is a different viewpoint: the question is equivalent to asking for a permutation $\sigma$ whose disjoint cycle structure contains exactly $m$ cycles, and with $|C_{S_{n}}(\sigma)|$ as small as possible subject to that.

Note that if $\sigma$ has $a_{j}$ cycles of length $j$ for each $j$ in that decomposition, then $|C_{S_{n}}(\sigma)| = \prod_{j} j^{a_{j}} a_{j}!,$ (which is just the denominator in the expression given in the question for the size of the conjugacy class). Note that this is clearly at least as large as $ \prod_{ \{j : a_{j} >0\}} j$, and note on the other hand that equality holds if all cycle sizes are distinct.

From now on I consider the case where there is a partition of $n$ into $m$ parts of distinct sizes (which is equivalent to requiring that $n \geq \frac{m(m+1)}{2}).$ I want to show first that there is a unique (up to conjugacy) choice of $\sigma$ which minimizes $|C_{S_{n}}(\sigma)|$ subject to the condition that $\sigma$ has $m$ cycles of pairwise distinct sizes.

Given such a permutation $\sigma,$ choose $j = j(\sigma)$ maximal subject to the fact that the $r$-th shortest cycle of $\sigma$ has size $r$ whenever $r \leq j$ - the possibility that $j =0$ is allowed, and occurs precisely when the shortest cycle of $\sigma$ has size greater than one. Note that the $j+1$-st shortest cycle of $\sigma$ (if there is one) has size at least $j+2$ by the maximality of $j$.

If $\sigma$ has more than $j+1$ disjoint cycles, then we may produce a permutation $\tau$ with $m$ disjoint cycles of pairwise distinct lengths and with $|C_{S_{n}}(\tau)| < |C_{S_{n}}(\sigma)|.$ We simply shorten the $j+1$-st shortest cycle of $\sigma$ by one, and lengthen the longest cycle of $\sigma$ by one. The resulting permutation $\tau$ still has all its $m$ disjoint cycles of pairwise distinct lengths, but we see easily that if $a$ is the length of the $j+1$-st shortest cycle of $\sigma$ and $b > a $ is the length of the longest cycle of $\sigma,$ then $ab|C_{S_{n}}(\tau)|= (b+1)(a-1)|C_{S_{n}}(\sigma)|$, so the claim follows as $(b+1)(a-1) < ab.$

Hence we can "improve" our choice of $\sigma$ if $\sigma$ has more than one cycle of length greater than $j+1,$ so the unique minimizing choice of $\sigma$ has its $r$-th cycle of length $r$ for $r \leq j$ and only one cycle of length greater than $j+1$. This forces $j = m-1$ or $j =m.$

Hence if $n \geq \frac{m(m+1)}{2},$ the unique choice of $\sigma$ with $m$ disjoint cycles of pairwise distinct lengths which minimizes $|C_{S_{n}}(\sigma)|$ has its $r$-th cycle of length $r$ for $1 \leq r \leq m-1$ and one cycle of length $n - \frac{m(m-1)}{2}.$ Note that the size of the corresponding conjugacy class is greater than $\frac{(n-1)!}{(m-1!)}$.

While this does not directly answer the question, it does perhaps give some insight.

Here is a different viewpoint: the question is equivalent to asking for a permutation $\sigma$ whose disjoint cycle structure contains exactly $m$ cycles, and with $|C_{S_{n}}(\sigma)|$ as small as possible subject to that.

Note that if $\sigma$ has $a_{j}$ cycles of length $j$ for each $j$ in that decomposition, then $|C_{S_{n}}(\sigma)| = \prod_{j} j^{a_{j}} a_{j}!,$ (which is just the denominator in the expression given in the question for the size of the conjugacy class). Note that this is clearly at least as large as $ \prod_{ \{j : a_{j} >0\}} j$, and note on the other hand that equality holds if all cycle sizes are distinct.

From now on I consider the case where there is a partition of $n$ into $m$ parts of distinct sizes (which is equivalent to requiring that $n \geq \frac{m(m+1)}{2}).$ I want to show first that there is a unique (up to conjugacy) choice of $\sigma$ which minimizes $|C_{S_{n}}(\sigma)|$ subject to the condition that $\sigma$ has $m$ cycles of pairwise distinct sizes.

Given such a permutation $\sigma,$ choose $j = j(\sigma)$ maximal subject to the fact that the $r$-th shortest cycle of $\sigma$ has size $r$ whenever $r \leq j$ - the possibility that $j =0$ is allowed, and occurs precisely when the shortest cycle of $\sigma$ has size greater than one. Note that the $j+1$-st shortest cycle of $\sigma$ (if there is one) has size at least $j+2$ by the maximality of $j$.

If $\sigma$ has more than $j+1$ disjoint cycles, then we may produce a permutation $\tau$ with $m$ disjoint cycles of pairwise distinct lengths and with $|C_{S_{n}}(\tau)| < |C_{S_{n}}(\sigma)|.$ We simply shorten the $j+1$-st shortest cycle of $\sigma$ by one, and lengthen the longest cycle of $\sigma$ by one. The resulting permutation $\tau$ still has all its $m$ disjoint cycles of pairwise distinct lengths, but we see easily that if $a$ is the length of the $j+1$-st shortest cycle of $\sigma$ and $b > a $ is the length of the longest cycle of $\sigma,$ then $ab|C_{S_{n}}(\tau)|= (b+1)(a-1)|C_{S_{n}}(\sigma)|$, so the claim follows as $(b+1)(a-1) < ab.$

Hence we can "improve" our choice of $\sigma$ if $\sigma$ has more than one cycle of length greater than $j+1,$ so the unique minimizing choice of $\sigma$ has its $r$-th cycle of length $r$ for $r \leq j$ and only one cycle of length greater than $j+1$. This forces $j = m-1$ or $j =m.$

Hence if $n \geq \frac{m(m+1)}{2},$ the unique choice of $\sigma$ with $m$ disjoint cycles of pairwise distinct lengths which minimizes $|C_{S_{n}}(\sigma)|$ has its $r$-th cycle of length $r$ for $1 \leq r \leq m-1$ and one cycle of length $n - \frac{m(m-1)}{2}.$ Note that the size of the corresponding conjugacy class is $\frac{n!}{(n-\frac{m(m-1}{2})(m-1)!}$.

While this does not directly answer the question, it does perhaps give some insight and provides large conjugacy classes of elements with $m$ disjoint cycles.