For symmetric group conjugacy classes and irreducible representation both are parametrized by Young diagramms, so there is a kind of "good" bijection between the two sets. For general finite groups see MO discussion.
Question: What are the finite groups where some "good" bijection(s) between conjugacy classes and irreducible representations are known ?
"Good" bijection is an informal "definition", nevertheless I hope example of S_n and other examples listed below, may convince that the question makes sense.
I think that it is far too optimistic to have one unique bijection for general group,
but it seems to me that for certain classes of groups there can be some set of "good" bijections. Let me briefly discuss below some properties which "good" bijection may satisfy, and may be discuss details in the next question.
Some examples:
1) symmetric group S_n
2) Z/2Z is naturally isomprhic to its dual, as well as $Z/2Z \oplus Z/2Z$ see e.g. MO "fantastic properties of Z/2Z"
3) Generally for abelian finite groups: among all set-theoretic bijections $G \to \hat G$, some are distinguished that they are group isomorphisms. So we have non unique, but
a class of "good" bijections.
4) For GL(2,F_q) Paul Garret writes: "conjugacy classes match in an ad hoc fashion with specic representations". (See here table at page 11).
5) G. Kuperberg describes relation of the McKay correspondence and that kind of bijection for A_5 (or its central extension), see here.
6) If I understand correctly here at MO D. Jordan mentions that bijection exists
for Coxeter groups. (I would be thankful for detailed reference).
7) Dihedral groups $D_{2n}$ – see answer by Glasby below
8) Finite Heisenberg group with $p^{2n+1}$ elements, also known as extraspecial group – see answer by Glasby below
9) Quternionic group $Q_{8}$ – this actually can be seen as a particular example of the item above. Or note that it is $Z/2Z$ central extension of $Z/2Z \oplus Z/2Z$,
and $ Z/2Z \oplus Z/2Z$ has natural bijection
as mentioned in item 2 above, and it is easy to extend it to $Q_8$.
10)
It seems that for Drinfeld double of a finite group (and probably more generally for "modular categories") there is known some analog of natural bijection.
There is such remark at page 5 of
Drinfeld Doubles for Finite Subgroups
of SU(2) and SU(3) Lie Groups.
R. COQUEREAUX, Jean-Bernard ZUBER:
In other words, there is not only an equal number of classes and
irreps in a double, there is also a canonical bijection between them.
There can be several properties which "good" bijection may satisfy, at least for some "good" groups
1) Respect the action of $Out(G)$. Actually the two sets are not isomorphic in general
as $Out(G)$-sets (see MO21606), however there are many cases where they are isomorphic see G. Robinson's MO-answer.
2) Reality/Rationality constraints. Again in general there is no correspondence see MO, but there are some cases where corresponding properties of classes and irreps agree – see J. Schmidt's answer on that question.
Two properties below are even more problematic
3) It might be that product on conjugacy classes have something to do with tensor product of representations (at least for abelian groups we may require these two fully agree).
4) If to think about kind of "orbit method" ideology, and think that conjugacy class is in some sense perversed coadjoint orbit, we may hope that structure of conjugacy representation,
should somehow respect the "good" bijection. For example for $S_n$ we proved that
irrep corresponding to Young diagram "d" lives inside the conjugacy subrepresentation
realized as functions on the conjugacy class parametrized exactly by "d".
(See MO153561, MO153991 for some discussion of conjugacy (adjoint) representation).
5) For algebraic groups over finite fields conjugacy classes and irreps
sometimes naturally divided into families (e.g. conjugacy classes are often
parametrized by equations $ F_{t_i}(x_k) = 0 $ – changing "t"
we get different conjugacy classes in the same "family").
So we may hope that good bijection respects the families.
(It is works fine for Heisenberg group, but for UT(4,p) I have met some problems).
Best Answer
This is an interesting question, even though it is not well defined. Call a group "good" if it has a "good" bijection between its conjugacy classes and its irreducible complex representations. I agree with Alexander that the definition of a "good" bijection/group should be guided by classes of examples, but I prefer that a class of bijections/groups should be infinite.
There are families of good metacyclic groups. For example, if $n=2m+1$ is an odd integer, then the dihedral groups $D_{2n}=\langle a,b\mid a^2=b^n=1,\; a^{-1}ba=b^{-1}\rangle$ of order $2n$ are good. The conjugacy classes $\{b^j,b^{-j}\}$, $1\leq j\leq m$, $\{1\}$, and $\{a,ab,ab^2,\dots,ab^{n-1}\}$ correspond bijectively (I believe this is "good") to the irreducible representations $\rho_j$, $1\leq j\leq m$, $\sigma_0$, and $\sigma_1$, respectively where $\rho_j(a)=\begin{pmatrix}0&1\\1&0\end{pmatrix}$, $\rho_j(b)=\begin{pmatrix}\zeta_n^j&0\\0&\zeta_n^{-j}\end{pmatrix}$, $\sigma_k(a)=\begin{pmatrix}(-1)^k\end{pmatrix}$, $\sigma_k(b)=\begin{pmatrix}1\end{pmatrix}$ and $\zeta_n=e^{2\pi i/n}$.
If an infinite family $G_1, G_2,\dots$ of groups is good, then you know a vast amount about each $G_n$ and can likely produce is a formula writing $|G_n|$ as a sum of the squares of the degrees of the irreducible representations. For $D_{2n}$ this is $2n=4m+2=m\times 2^2+2\times 1^2$. If $G_n$ is an extraspecial group of (odd) order $p^{1+2n}$ and exponent $p$, then the formula is $p^{1+2n}=(p-1)\times(p^n)^2+p^{2n}\times 1^2$. Perhaps the existence of such a formula should be part of the elusive definition of "good".
Addition: Yes Alexander, you are correct, the extraspecial groups $G_n$ of order $p^{1+2n}$ and odd exponent $p$ are "good". To describe a "good" bijection I need some notation. Let $f_n\colon V\times V\to\mathbb{F}_p$ be a nondegenerate symplectic form on $V=\mathbb{F}_p^{2n}$. Multiplication in $G_n=V\times \mathbb{F}_p$ is given by $(v_1,\lambda_1)(v_2,\lambda_2)=(v_1+v_2,\lambda_1+\lambda_2+{\frac12}f_n(v_1,v_2))$, or by the matrices you indicate. The conjugacy classes are as follows: the $p$ one-element (central) classes $Z_\lambda:=\{(0,\lambda)\}$, $\lambda\in\mathbb{F}_p$, and the $p^{2n}-1$ classes $C_v:=\{(v,\lambda)\mid \lambda\in\mathbb{F}_p\}$ where $0\neq v\in V$. The irreducible representations are also easy. The trivial degree-1 representation corresponds to class $Z_0$ containing the identity element. The $p^{2n}-1$ nontrivial degree-1 representations correspond to the $p$-element classes $C_v$. The remaining $p-1$ irreducibles of degree $p^n$ correspond to the $p-1$ central conjugacy classes $Z_\lambda$, with $0\neq\lambda\in\mathbb{F}_p$. Fix a maximal totally isotropic subspace $W$ of $V$. By Witt's theorem $|W|=p^n$. Then $A:=W\times\mathbb{F}_p$ is a maximal abelian subgroup of $G_n$ of index $p^n$. Let $\sigma_\lambda$ be the 1-dimensional representation of $A$, with kernel $W$, mapping $(0,1)$ to $e^{2\pi i\lambda/p}$. The induced representations $\rho_\lambda={\rm Ind}_A^{G_n}(\sigma_\lambda)$ are irreducible of degree $p^n$. (A direct calculation shows $\langle\rho_\lambda,\rho_\lambda\rangle=1$. Choosing a different $f_n$, or a different maximal totally isotropic subspace $W'$, gives equivalent representations $\rho'_\lambda$. The $W$s are permuted by ${\rm Aut}(G_n)$.) This is a "good" bijection, as identifying $V$ with its dual seems allowed.