Is there some natural bijection between irreducible representations and conjugacy classes of finite groups (as in case of $S_n$)?
Group Theory – Bijection Between Irreducible Representations and Conjugacy Classes of Finite Groups
bijective-combinatoricsfinite-groupsgr.group-theoryrt.representation-theory
Related Solutions
The answer to your first question is negative. For a concrete example, you can show that the conjugacy class rings of the nonisomorphic groups $Q_8$ and $D_8$ are isomorphic, via an isomorphism that pairs off the bases as follows: $[1] \leftrightarrow [1]$, $[-1] \leftrightarrow [r^2]$, $[i] \leftrightarrow [r]$, $[j] \leftrightarrow [s]$ and $[k] \leftrightarrow [rs]$.
As to your question about the relationship between the conjugacy class ring and the character ring, there are lots of partial results that can be stated. Nonetheless, the answer to the question of when these two rings are isomorphic is completely known. This turns out to be the case if and only if the group is $p$-nilpotent with abelian Sylow $p$-subgroup. More generally, for arbitrary finite groups $G$ and $G'$, the following two conditions are equivalent.
The character ring of $G$ is isomorphic to the conjugacy class ring of $G'$.
$G$ and $G'$ are $p$-nilpotent groups with abelian Sylow $p$-subgroups. Moreover, if $g_1, \dots, g_l$ and $g_1',\ldots, g_{l'}'$ are complete sets of representatives for the conjugacy classes of $p'$-elements of $G$ and $G'$, resp., and if $D_i$ and $D_i'$ are Sylow $p$-subgroups of $C_G(g_i)$ and $C_{G'}(g_i')$, resp., then $l=l'$ and $D_i \cong D_i'$.
This is due to Saksonov, The ring of classes and the ring of characters of a finite group. Mat. Zametki 26 (1979), no. 1, 3–14, 156.
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.
Best Answer
This is a different take on Steven Landsburg's answer. The short version is that conjugacy classes and irreducible representations should be thought of as being dual to each other.
Fix an algebraically closed field $k$ of characteristic not dividing the order of our finite group $G$. The group algebra $k[G]$ is a finite-dimensional Hopf algebra, so its dual is also a finite-dimensional Hopf algebra of the same dimension; it is the Hopf algebra of functions $G \to k$, which I will denote by $C(G)$. (The former is cocommutative but not commutative in general, while the latter is commutative but not cocommutative in general.) The dual pairing $$k[G] \times C(G) \to k$$
is equivariant with respect to the action of $G$ by conjugation, and it restricts to a dual pairing $$Z(k[G]) \times C_{\text{cl}}(G) \to k$$
on the subalgebras fixed by conjugation; $Z(k[G])$ is the center of $k[G]$ and $C_{\text{cl}}(G)$ is the space of class functions $G \to k$. Now:
The second identification should be clear; the first comes from considering the central character of an irreducible representation. Now, the pairing above is nondegenerate, so to every point of the maximal spectrum of $Z(k[G])$ we can canonically associate an element of $C_{\text{cl}}(G)$ (the corresponding irreducible character) and to every point of the maximal spectrum of $C_{\text{cl}}(G)$ we can canonically associate an element of $Z(k[G])$ (the corresponding sum over a conjugacy class divided by its size).