Centralizer of a cyclic subgroup within the group algebra $\mathbb{C} S_N$ of the symmetric group Let us take the group algebra $\mathbb{C} S_N$ and the subgroup $H=Z_N$ generated by the element
$\sigma=(123\dots N)$, which is a cyclic shift. What is the structure of the centralizer of $H$ within $\mathbb{C} S_N$? If we
are looking at the Lie algebra $\mathcal{L}(\mathbb{C} S_N)$ obtained from the usual 
commutator of two elements, then what is the centralizer of the commutative sub-algebra $\mathbb{C} H$?
My motivation for this comes from physics, where the special ordering $1,2,3,\dots N$ has a
concrete physical meaning, for example actual atoms are placed next to each other. In this case those
group algebra elements that commute with $\sigma$ are called ``translationally invariant''
permutation operators, under periodic boundary conditions.
Ultimately I would like to know what is the maximal number of mutually commuting operators within
$\mathbb{C} S_N$ that commute with $\mathbb{C} H$, and how to find explicit bases for them. I wanted to approach this through the structure of
the Lie algebra $\mathcal{L}(\mathbb{C} S_N)$, decomposition into simple Lie algebras, etc. I understand
that the irreducible subspaces in $\mathbb{C} S_N$ correspond to the Young symmetrizers. But it is not
so clear what to do from here, how to add the action of the concrete element $\sigma$ into this.
 A: Thanks to Mark Wildon and the OP for pointing out that my answer was incorrect- I was considering the centralizer of $\sigma$ in the wrong algebra.
    However, it does seem to me that the structure of $A = C_{\mathbb{C}S_{N}}(\sigma)$ depends on the prime factorization of $N.$
In general it is well-known that if $\lambda$ is a partition of $N$ and $\chi_{\lambda}$ is the associated complex irreducible character of $S_{N}$, then $\chi_{\lambda}(\sigma) \in \{0,1,-1 \}$.
When $N= p$ is prime, this implies that ${\rm Res}^{S_{N}}_{\langle \sigma \rangle}(\chi_{\lambda})$ has the form $t\rho + \chi_{\lambda}(\sigma)1, $ where $t= \frac{\chi_{\lambda}(1) - \chi_{\lambda}(\sigma)}{p}$ is a non-negative integer and $\rho$ is the regular character of $\sigma$. But this is never the case for all $\chi_{\lambda}$ when $N$ is not prime. 
Now $\mathbb{C}S_{N}$ is isomorphic to $\bigoplus_{\lambda} M_{\chi_{\lambda}(1)}( \mathbb{C})$ as $\lambda$ runs through partitions of $N$. Now $\sigma$ acts as a matrix of trace $0$ or $\pm 1$ inside  $M_{\chi_{\lambda}(1)}( \mathbb{C}).$
In the former case, the fixed subalgebra of $\sigma$ on the matrix algebra $M_{\chi_{\lambda}(1)}( \mathbb{C})$ has dimension
$\frac{\chi_{\lambda}(1)^{2}}{p}.$
In the latter cases, we may compute the dimension of the fixed fixed subalgabra of $\sigma$  in the relevant matrix algebra.
If ${\rm Res}^{S_{N}}_{\langle \sigma \rangle }(\chi_{\lambda})= t_{\lambda} \rho  \pm 1,$ then the fixed subalgebra of $\sigma$ in the matrix algebra 
has dimension $(p-1)t_{\lambda}^{2} + (t_{\lambda} \pm 1)^{2}
= pt_{\lambda}^{2} \pm 2t_{\lambda} +1$.
This means that when $N = p$ is prime, the dimension of the centralizer algebra of $\sigma$ in $\mathbb{C}S_{N}$ is totally detrmined by the values of the $\chi_{\lambda}(1)$ and $\chi_{\lambda}(\sigma)$, but his is not the case when $N$ is not prime.
A: $\newcommand{\IC}{\mathbb{C}}$
Let $V_\lambda$ be the irreducible $S_N$-module (Specht module). The representations $\rho_\lambda: \IC S_N \to \operatorname{End}_\IC(V_\lambda)$ give us an isomorphism of algebras $\IC[S_N] \to \prod_{\lambda \vdash N} \operatorname{End}_\IC(V_\lambda)$.
Since we work with $\IC$, the endomorphisms of finite order are always diagonalisable. In particular we can decompose $V_\lambda = \bigoplus_{k=0}^{N-1} V_{\lambda,k}$ such that $\sigma$ acts as multiplication by $\exp(\frac{2\pi i}{N}k)$ on $V_{\lambda,k}$. The centraliser of $\sigma$ is therefore exactly equal to $\prod_{\lambda\vdash N, 0\leq k<N} \operatorname{End}_\IC(V_{\lambda,k})$.
This reduces the problem to finding the commutative subalgebras of $\IC^{d\times d}$ of maximal dimension for all dimensions $d$. If I'm not mistaken, the subalgebra of diagonal matrices is of maximal dimension among the commutative subalgebras of $\IC^{d\times d}$, i.e. the maximal dimension is just $d$ itself.
Therefore the maximal dimension of a commutative subalgebra inside the $C_{\IC S_N}(\sigma)$ is $\sum_{\lambda,k} \dim(V_{\lambda,k}) = \sum_\lambda \dim(V_\lambda)$. The dimensions $\dim(V_\lambda)$ can be calculated by the hook length formula.
In principle this approach also tells you how to find a set of basis elements: Find an eigen-basis for $\sigma$ inside each $V_\lambda$. The diagonal matrices w.r.t. to this basis will give you a basis of a commutative subalgebra of maximal dimension.
