Let $\lambda$ be a partition of $n \in \mathbb Z_{>0}$. Then $\lambda$ gives rise to a Young diagram. The natural action of the symmetric group $S_n$ gives rise to Young tabloids, which we order using the lexicographic ordering. There will be $[S_n/ S_\lambda]$ Young tabloids of shape $\lambda$, where $S_\lambda$ is the stabiliser of $\lambda$ under the action of $S_n$.
This can be phrased more simply too: for example, for the partition $(2, 2)$ of $4$, I want to consider $(a, a, b, b) > (a, b, a, b) > (a, b, b, a) > (b, a, a, b) > (b, a, b, a) > (b, b, a, a)$. Note that there are $6 = [S_4 / (S_2 \times S_2)]$ objects -- call them $Y_1, \dots, Y_6$.
The action of $S_r$ on $[n] = \{1, \dots, n\}$ induces an action on the set $Y(\lambda)$ of Young tabloids of shape $\lambda$. Identify $Y(\lambda)$ as an ordered set with the set of natural numbers between $1$ and $[S_r/S_\lambda]$.
Going back to the example, let $\sigma = (1 2)$. Then $\sigma Y_1 = Y_1, \sigma Y_2 = Y_5, \sigma Y_3 = Y_6, \sigma Y_4 = Y_4, \sigma Y_5 = Y_2, \sigma Y_6 = Y_3$. Thus $\sigma$ acts as $(2 5)(3 6)$.
Another example: $\lambda = (1, 1, 1)$ and $\sigma = (1 2 3)$. Then $\sigma$ acts on $Y(\lambda)$ through $(1 4 5)(2 6 3)$.
Let $\sigma \in S_n$. What is the relationship between the cycle $\sigma$ acting through the regular representation on $[n]$ and $\sigma$ acting on $Y(\lambda)$ as above, also given as a cycle?
I hope I explained my problem adequately, I'm new to Young diagrams and the like. I'd be grateful for hints as well.
The answer is more complicated than a simple formula. I will outline a very rough procedure.
Suppose $\sigma\in S_n$ is a permutation and its action on $Y(\lambda)$ has cycle type $1^{c_1}2^{c_2}\cdots n^{c_n}$. Then for any $1\le k\le n$ we have the fixed point count
$$ |Y(\lambda)^{\large\sigma^{\large k}}| = \sum_{d\mid k} c_d $$
By Mobius inversion this yields the cycle count
$$ c_k=\sum_{d\mid k}\mu\big(\tfrac{k}{d}\big)|Y(\lambda)^{\large \sigma^{\large d}}|. $$
Any $\ell$-cycle of $\sigma$ becomes $(d,\ell)$-many $\ell/(d,\ell)$-cycles in $\pi=\sigma^d$.
Say $\pi$ has cycle type $1^{b_1}2^{b_2}\cdots n^{b_n}$ acting on $Y(\lambda)$, so
$$ b_k=\sum_{k=\ell/(d,\ell)} (d,\ell)c_\ell. $$
Finally, the fixed point count is given directly by
$$ |Y(\lambda)^\pi|=\sum_{\substack{\sum_{\large i} im_{ij}=\lambda_j \\ \sum_{\large j} m_{ij}=b_i}} \prod_i \binom{b_i}{m_{i1},m_{i2},\cdots}. $$
The matrix $M$ dictates the sizes of cycles whose supports are unioned together to get the rows of a tabloid, and the multinomial coefficients corresponding to choosing exactly which cycles of given sizes are used in which rows.