"Conjugacy classes–irreducibles" bijection, but for permutation representations

Question

The linear representation theory (over say $\mathbb{C}$ for concreteness) of a finite group $G$ is "the same as" its character theory. Characters are naturally functions on conjugacy classes of elements ("class functions"), and the number of irreducible representations is the same as the number of conjugacy classes of elements. But in general there is no canonical bijection between conjugacy classes of elements and irreducible representations.

Now let's consider permutation representations, i.e., $G$-sets. The analog of characters for permutation representations are "marks". Marks are naturally functions on conjugacy classes of subgroups of $G$, and the number of "irreducible" $G$-sets is the same as the number of conjugacy classes of subgroups of $G$. But now we do have a canonical bijection between conjugacy classes of subgroups and irreducible $G$-sets: each conjugacy class $H$ naturally determines an orbit $G/H$.

Question: Is there a high-level explanation for this apparent "difference" between the linear representation theory and the permutation representation theory of finite groups?

Answer 1

The difference is apparent, but not perhaps real. We are not saying that there isn't a 'canonical' basis for the set of class functions, just that it isn't the irreducible characters.

Here is an arguable definition of a canonical basis for the set of class functions. Let $G$ be finite, and $x\in G$ of order $n$. Define $\phi_x$ to be the character of $\langle x\rangle$ that maps $x$ to $\mathrm{e}^{2\pi \mathrm{i}/n}$, and let $\chi_x$ be the induction of $\phi_x$ to $G$. Notice that if $x$ and $y$ are conjugate in $G$ then $\chi_x=\chi_y$, and indeed the $\chi_x$ form a basis for the set of class functions.

That's pretty natural a basis to choose, it is just far from the irreducible characters in general. I don't know if this answers your question.