Eigenvalues of operator on $S_n$'s group algebra Take the group algebra of the symmetric group $S_n$ (or equivalently consider $S_n$'s regular representation) - I guess over $\mathbb{C}$.
If $e_{i,j} \in S_n$ denotes the element which swaps only the pair $i$ and $j$, consider the linear operator
$$
O = (e_{1,2} + e_{2, 3} + \cdots + e_{n-1, n} + e_{n,1})
$$
which acts on the group algebra.
The question is, for generic $n$, what are the eigenvalues of this operator?
For example, in the easy case of $S_2$, the operator $O$ is $e_{1,2} + e_{2,1} = 2e_{1, 2}$, the group algebra is two-dimensional, and $O$ has eigenvectors $1_{S_2} + e_{1, 2}$ and $1_{S_2} - e_{1, 2}$ with eigenvalues $2, -2$ respectively.
 A: For $n=4$, the eigenvalues are $4, -4, 0, 0, 0, 0, 0, 0, 2, 2, 2, 2, 2, 2, 2, 2, -2, -2, -2, -2, -2, -2, -2, -2$.
For $n=5$, the eigenvalues are the roots of
$(x - 5) \cdot (x + 5) \cdot (x - 1)^5 \cdot (x + 1)^5 \cdot x^{24} \cdot (x^2 - 5)^6 \cdot (x^2 - 5\cdot x + 5)^8 \cdot (x^2 + 5\cdot x + 5)^8 \cdot (x^2 - 2\cdot x - 4)^{10} \cdot (x^2 + 2\cdot x - 4)^{10}$
(courtesy of Sage cell server), so don't get your hopes up for rational spectra. These are not the Young-Jucys-Murphy elements.
For the sake of reference, here is my crappy Sage code:
n = 5
QSn = SymmetricGroupAlgebra(QQ, n)
sn = list(Permutations(n))
eln = QSn(Permutation([n]+range(2, n)+[1]))
for i in range(1, n):
    eln += QSn(Permutation(range(1, i) + [i+1, i] + range(i+2, n+1)))
mx = []
for p in sn:
    elf = eln * QSn(p)
    mx.append([elf.coefficient(q) for q in sn])
m = matrix(QQ, mx)
print m.characteristic_polynomial().factor()

For $n=6$, degree-$3$ factors appear, so don't hope for anything semi-nice either.
A: The eigenvalues are symmetric about 0, since for $n>2$, the Cayley graph $\Gamma_n$ of $S_n$ over the set $\{(1,2),\dots,(n-1,n),(1,n)\}$ is bipartite (consider odd permutations and even permutations as bipartition sets) and the spectrum of $\Gamma_n$ is  the spectrum of the group algebra element. Note that the spectrum of a bipartite graph is symmetric about 0.
