If $G$ is a finite group and its conjugacy classes are known, can the conjugacy classes of the wreath product $G \wr S_n \cong G^n \rtimes S_n$ be determined?

4$\begingroup$ Yes, this is well known. It is described in detail in James and Kerber's book on the representation theory of symmetric groups. $\endgroup$ – Christian Gaetz Oct 9 '18 at 15:59

2$\begingroup$ Did you mean $G\wr S_n\cong G^n\rtimes S_n$? $\endgroup$ – Alex B. Oct 9 '18 at 16:14

$\begingroup$ Alex: I mean for $S_n$ to be acting on $n$ copies of $G$, in the natural way. Have I gotten the notation the wrong way round? $\endgroup$ – Chris Russell Oct 9 '18 at 16:19

1$\begingroup$ In that case Chris, the notation is the wrong way round, and Alex's is what you meant ( and I think the group dealt with in JamesKerber as mentioned by Christian). $\endgroup$ – Geoff Robinson Oct 9 '18 at 16:25
This is indeed handled in Section 4.2 of James and Kerber's book on Representations of the Symmetric group.
We also considered this problem in Section 2 our paper describing a computer algorithm for computing conjugacy class representatives in permutation groups.
J. Cannon and D. Holt, Computing conjugacy class representatives in permutation groups, J. Algebra 300 (2006), 213222.
I can cut and paste the main bit of theory you need. The problem easily reduces to finding those classes that map onto a specific element of $S_n$.
Let $A$ be any group, let $P$ be a permutation group acting on the set $\{ 1\ldots d\}$, and let $W := A \wr P$. Then $W$ is a semidirect product of $A^d$ by $P$, and elements of $W$ have the form $(g,x)$ with $g \in P$, $x \in A^d$, and $x = (x_1,\ldots,x_d)$ with $x_i \in A$. In general, for $x \in A^d$, we denote the $i$th component of $x$ by $x_i$. The action of $P$ on $A^d$ in $W$ is given by: $$(g,1)^{1}(1,(x_1,\ldots,x_d))(g,1) = (1,(x_{1^{g^{1}}},\ldots, x_{d^{g^{1}}})).$$
Hence, for $g \in P$, $x,z\in A^d$, we have \begin{equation}\label{conjeqn} (1,z)^{1}(g,x)(1,z) = (g,y), \ \mathrm{where}\ y_i = z^{1}_{i^{g^{1}}}x_iz_i \ \mathrm{for}\ 1 \le i \le d. \end{equation}
$\textbf{Theorem}$ With the above notation, let $x^{(1)},\ldots,x^{(k)}$ be representatives of the conjugacy classes of $A$. Fix an element $g \in P$, and let $r_1,r_2,\ldots,r_s$ be representatives of the cycles of $g$ in its action on $\{ 1\ldots d\}$. Then $$ R := \{\,(g,x) \mid x_i = 1 \ \mathrm{for}\ i \not\in \{r_1,\ldots,r_s\},\, x_i \in \{x^{(1)},\ldots,x^{(k)}\}\ \mathrm{for}\ i \in \{r_1,\ldots,r_s\}\,\} $$ is a set of representatives of the $A^d$classes of elements of $W$ of the form $(g,x)$ for $x \in A^d$.
The size of the $A^d$class $R_{gx}$ containing $(g,x)$ is the product of the sizes of the classes of the $x_i$ in $A$ with $i \in \{r_1,\ldots,r_d\}$ with the lengths of the cycles of $g$ that contain $r_1,\ldots,r_d$.
The above describes the $A^d$classes of $W$; i.e. the orbits under conjugation by the subgroup $A^d$ of $W$. To get the classes of $W$ itself, first of all we only consider those $R_{gx}$ as $g$ ranges over a set of class representatives of $P$. Then, for each such $g \in W$, the $A^d$class representatives $R_{gx}$ are fused under the action of $C_P(g)$. This action is the same as the action of $C_P(g)$ on the cycles of $g$, so is easy to calculate.
Then the size of a $W$class $R_{gx}$ is the product of the size of the class of $g$ in $P$ with the sum of the the sizes of the $A^d$classes $R_{gx'}$ that lie in the orbit under the action of $C_P(g)$.

$\begingroup$ Thank you Derek, I see how an element $(g, x)$ is conjugate to something in the set $R$ you define. If I write $$R_g := \{\,(g,x) \mid x_i = 1 \ \mathrm{for}\ i \not\in \{r_1,\ldots,r_s\},\, x_i \in \{x^{(1)},\ldots,x^{(k)}\}\ \mathrm{for}\ i \in \{r_1,\ldots,r_s\}\,\}$$ is there a way to describe a subset of $$\bigcup_{g \in G}R_g$$ which contains only one representative from each conjugacy class? Also is there a way to determine how many elements are in a each conjugacy class? $\endgroup$ – Chris Russell Oct 10 '18 at 9:32

1$\begingroup$ My answer was incomplete because it only described the $A^d$classes. Calculating the $G$classes from thsi information is not difficult  I have added some further explanation. $\endgroup$ – Derek Holt Oct 10 '18 at 10:48

$\begingroup$ I think I get the idea of how it works now but I'm not sure what $G$ is in the second last paragraph of your updated answer, which is confusing me. In one place I'm interpreting that $A^d$ is a subgroup of $G$ but in another $g \in G$ where I thought $g$ is an element of $P$. $\endgroup$ – Chris Russell Oct 10 '18 at 11:29

$\begingroup$ Sorry $G$ should be $W$. I don't have a $G$. I have $W = A^d \rtimes P$. Of course $P=S_n$ in your question. $\endgroup$ – Derek Holt Oct 10 '18 at 12:10