For any group character $\chi: G \to \mathbb{C}$, the function $\chi(g^k)$ is an integer-linear combination of other group characters in $\mathrm{Irr}(G)$. There doesn't seem to be a general way of finding the expansion in terms of the basic representations.

In the case of the symmetric group $S_n$, the permutation representation (dim = n ) decomposes into the standard representation (dim = n-1) and the trivial representation (dim = 1). $\mathrm{perm} = (n) + (n-1,1)$. I like this representation b/c it's character is the number of fixed points.

I would like to decompose $\chi_{\mathrm{perm}}(g^k)$ into an integer combination of irreducible characters. Fulton-Harris gives a formula for the exterior powers of the standard representation: $\wedge^d (n-1,1)= (n-d,d)$. I would settle for expanding $\chi_{(n-1,1)}(g^k)$ into irreducible characters.

In terms of eigenvalues these are the elementary symmetric polynomials $e_1 = \lambda_1 + \dots \lambda_n$ and $e_2 = \sum_{i < j} \lambda_i \lambda_j$ and I'm looking for the symmetric powers $\lambda_1^k + \dots + \lambda_n^k$. Maybe there's a way using Newton's identities, but that doesn't explain how to expand the tensor product into irreducibles.

Probably, $\chi_{\wedge^k \mathrm{perm}}(g)$ counts fixed k-element subsets of {1,2,...,n}, which is different than $\chi_{\mathrm{perm}}(g^k)$ counting (possibly degenerate) fixed $k$-cycles.