This question was inspired by Smith Normal Form of a Cayley Graph of the Symmetric Group. Let $\mathbb{Q}S_n$ denote the group algebra over $\mathbb{Q}$ of the symmetric group $S_n$. Identify a conjugacy class $K_\lambda$ (where $\lambda$ is a partition of $n$) with the sum of its elements in $\mathbb{Q}S_n$. Let $Z_n$ denote the center of $\mathbb{Q}S_n$, so a $\mathbb{Q}$-basis for $Z_n$ consists of the sums $K_\lambda$. $K_\lambda$ acts on $Z_n$ by left multiplication. What is the Smith normal form (SNF) over $\mathbb{Z}$ of the matrix of this action with respect to the basis $\{K_\mu\}$?

Of special interest is the case $\lambda=(n)$, so $K_{(n)}$ is the class (or sum) of $n$-cycles. There are $n$ nonzero diagonal elements of the SNF. (This is easy to see.) I have computed them up to $n=13$. The $k$th diagonal entry ($0\leq k\leq n-1$) is $k!$ times a rational number with small numerator and denominator. The largest diagonal entry is always $(n-1)!$. For instance, when $n=9$ the SNF is $$ 0!,\ 2\cdot 1!,\ 2!,\ \frac 23 3!,\ 4!,\ 2\cdot 5!,\ \frac 136!,\ 2\cdot 7!,\ 8!. $$ When $n=12$ the SNF is $$ 0!,\ 1!,\ 2!,\ \frac 133!,\ \frac 124!,\ 5!,\ 2\cdot 6!,\ 7!,\ \frac 128!,\ \frac 139!,\ 10!,\ 11!. $$

Two specific conjectures:

(a) If $n$ is an odd prime, then the nonzero SNF terms are $k!$ for $k$ even and $2\cdot k!$ for $k$ odd, where as above $0\leq k\leq n-1$.

(b) If $n$ is twice an odd prime, then the nonzero SNF terms are $k!$ for all $0\leq k\leq n-1$, except that $(n/2)!$ is omitted, and $(\frac n2-1)!$ appears twice.

Possibly with more data one could make a conjecture for all $n$.

Note that my question makes sense for any finite group. Can anything be said about the SNF in this generality?