Can two different characters of $S_n$ have the same _multiset_ of values? As I was going through various representation-theory posts in the site, I stumbled upon this one: Characters of the symmetric group corresponding to partitions into two parts. Now, that question hasn't been answered but there is a link in the comments to the following paper by Mark Wildon, Labelling the character tables of symmetric and alternating groups (you can find the published version here).
In that paper, Mark shows that for $n\neq 4,6$ given an unlabelled character table of $S_n$, there is a unique way to label its rows and columns naturally (that is, in such a way that $\chi_{\lambda\nu}=\chi^{\lambda}(\nu)$). From what I understand, this says that the character table isn't that symmetric.
Now, this is definitely a beautiful result, but my question regards his comment after Proposition 2.3. The proposition gives examples of conjugacy classes whose multisets of character values are the same (so you cannot distinguish between them). After that he asks whether such examples exist for the rows of the character table too. In particular, he says this never happens for $n\leq 30$.
Can two different characters of $S_n$ have the same multiset of values? Does anything change if we consider reducible characters as well? What happens in arbitrary (nonabelian) groups $G$, are there easy counterexamples there?
 A: Fleshing out the comment a bit.
Assume that the group $S_n$ has two non-trivial conjugacy classes of the same size, say $|[\sigma]|=|[\tau]|=C$ for some non-conjugate permutations $\sigma,\tau\in S_n$ and some positive integer $C\mid n!$. Define two class function $\psi_1$ and $\psi_2$
by declaring that $\psi_1(1)=Kn!=\psi_2(1)$, $\psi_1(\tau)=n!/C=\psi_2(\sigma)$ and setting the remaining values of both to zero.
Let $\chi_1,\chi_2,\ldots,\chi_\ell$ be the irreducible characters of $S_n$. It is known that $\chi_i(\pi)$ is a rational integer for all $\pi\in S_n, i=1,2,\ldots,\ell$.
Let's calculate the inner products:
$$
(\chi_i\mid \psi_1)=\frac1{n!}\left[\chi_i(1)\psi_1(1)+C\cdot\chi_i(\tau)\psi_1(\tau)\right]=K\cdot\chi_i(1)+\chi_i(\tau),
$$
and
$$
(\chi_i\mid \psi_2)=\frac1{n!}\left[\chi_i(1)\psi_2(1)+C\cdot\chi_i(\tau)\psi_2(\sigma)\right]=K\cdot\chi_i(1)+\chi_i(\sigma).
$$
These inner products are thus automatically rational integers. Whenever $K$ is large enough, all these integers are non-negative (and at least one is positive). For such a choice of $K$ both class functions are thus characters.
Incrementing $K$ by one amounts to summing the regular representation to the virtual character.
Of course, both $\psi_1,\psi_2$ are both highly reducible. The point of this note was to make it clear that the question becomes quite easy, if we allow reducible characters.
