Let $\lambda \vdash nk$. Let $n^k$ denote the partiton with $k$ parts of size $n$. We can compute $\chi^\lambda(n^k)$ by using the [Murnaghan-Nakayama rule,](https://en.wikipedia.org/wiki/Murnaghan%E2%80%93Nakayama_rule) as a signed sum over border-strip tableaux, (shape $\lambda$ and border strips of size $n$). 


Now, by checking several cases on the computer, it seems like each term in the (Murnaghan-Nakayama) sum has the same sign. Combinatorially, it reduces to proving the following:


Define the height of a border-strip to be the maximal row index that it touches, minus the minimal row index. The height of a border-strip decomposition of the diagram $\lambda$ is the sum of the heights of the border-strips. Show that the parity of this sum is independent of the actual decomposition (it only depends on $\lambda$ and $n$).


I am looking for a reference of this (I do think I can in fact prove a stronger statement, but I have not worked out the details yet).



Murnaghan-Nakayama rule when all cycles have same size
This is corollary 10 in ["A bijection proving orthogonality of the characters of $S\_n$"](https://www.sciencedirect.com/science/article/pii/0001870883900385), by Dennis E. White.


