TITLE: Dimension of Specht Modules $S^\lambda$
QUESTION [7 upvotes]: In the study of representation theory of $S_n$, we know that the irreducible characters of $\chi_\lambda$ of $S_n$ are indexed by partitions $\lambda \vdash n$. There are several methods in determining the dimension of each $S^\lambda$, $f^\lambda$. One of the method is by considering
$ f^\lambda = \frac{n!}{\prod \text{hook length}} $
Let $\lambda' \vdash n$ be a partition obtained by taking 'transpose' in Ferrer's diagram of $\lambda$. For example, if $\lambda = (5,4,1)$, then $\lambda' = (3,2,2,2,1)$. Using the formula of $f^\lambda$ above, we thus have $f^\lambda = f^{\lambda'}$.
After some observations, I found out that when $n \geq 8$, then the dimension of $S^\lambda$ is unique up to transpose. In other words,
"Given any $\lambda \vdash n$, then there exists no other $\alpha \vdash n$ such that $f^\lambda = f^{\alpha}$ except when $\alpha = \lambda$. "
Is the above result well-known or established by anyone?
Thanks for the help!

REPLY [2 votes]: An even stronger counter-example is presented in Example 8 in my paper.
It shows that $\lambda = (8,5,4)$ and $\mu=(7,7,2,1)$
both have the same multi-set of hook values. This of course implies $f^\lambda=f^\mu$.
What is interesting though is that the corresponding Ehrhart functions for the posets defined by $\lambda$ and $\mu$
are different.
Perhaps you can salvage your conjecture, and ask if for $|\lambda|>8$, the Ehrhart polynomial is unique (up to transposition). The value of $f^\lambda$ can be recovered from the leading coefficient of the Ehrhart polynomial.