It is a well-known fact in a group representation theory that all elements of the same conjugacy class possess the same trace (character) in any representation of a group. I deal specifically with symmetric groups $S_n$, which are generated by basic generators $\sigma_{i,i+1}$ with $i=1...n-1$. Two elements of the symmetric group belong to the same conjugacy class if they represent permutations with the same cyclic structure. My question is the following: can one take some representation of the symmetric group (with $\sigma_{i,i+1}$ mapped to matrices $T_{i,i+1}$) and deform it (probably sacrificing the group structure) in such a way (basic generators $\sigma_{i,i+1}$ are mapped now to some matrices $\tilde{T}_{i,i+1}$) that traces of products of some prescribed amount of generators still depend only on the cyclic structure of the corresponding permutation?
One obvious example is to take matrices $T_{i,i+1}$ of any representation of $S_n$ and to multiply them by a number. If $\sigma_{i,i+1}$ is mapped to $\alpha T_{i,i+1}$ then trace of product of $m$ matrices will be equal to $\alpha ^m$ times a character of $S_n$ and will depend only on the cyclic structure of generators. A less trivial example is to take permutation matrices and to add to each matrix element the same constant: $(\tilde{T}_{i,i+1})_{pq}=(T_{i,i+1})_{pq}+a$, where $(T_{i,i+1})_{pq}$ are matrix elements of permutation matrices.
I wonder if there are more examples of such a behavior and if there is an underlying theory that allows one to construct such deformations.