The Alternating group $A_5$ on five elements has conjugacy classes very similar to those of the Symmetric group $S_5$, but where one of the classes has split in two.

Note that $A_5$ has $60$ elements, since it is precisely half of the elements of $S_5$ which has $5! = 120$ elements (where the exclamation mark is the Factorial function).

Table

$$\begin{array}{|c|c|c|c|} \hline \text{Representative}& \text{Size of class} & \text{Cycle type} & \text{Order of element} \\ \hline (12345) & 12 & 5 & 5 \\ \hline (21345) & 12 & 5 & 5 \\ \hline (123) & 20 & 3,1,1 & 3 \\ \hline (12)(34) & 15 & 2,2,1 & 2 \\ \hline e & 1 & 1,1,1,1,1 & 1 \\ \hline \end{array}$$

Working

Firstly, the identity is in a class of its own, because $\tau e \tau^{-1} = \tau \tau^{-1} = e$ for every $\tau$.

Now, by the same reasoning as in the proof that conjugate elements must have the same cycle type in $S_n$, that result also holds in $A_n$.

Hence we just need to see whether any of the cycle types comprise more than one conjugacy class.

Recall that the available cycle types are $(5)$, $(3,1,1)$, $(2,2,1)$, $(1,1,1,1,1)$ (the last of which is the identity and we have already considered it).

Double-transpositions

All the double-transpositions are conjugate (so the $(2,2,1)$ cycle type does not split):

• $(ab)(cd)$ is conjugate to $(ab)(ce)$ if we conjugate by $(ab)(de)$; symmetrically this covers all the cases where one of the two transpositions remains the same.
• $(ab)(cd)$ is conjugate to $(ac)(bd)$ by $(cba)$; this covers the case that $e$ is not introduced.
• $(ab)(cd)$ is conjugate to $(ac)(be)$ by $(bc)(de)$; this covers the remaining cases that $e$ is introduced and neither of the two transpositions remains the same.

Three-cycles

All the three-cycles are conjugate (so the $(3,1,1)$ cycle type does not split):

• $(abc)$ is conjugate to $(acb)$ by $(bc)(de)$, so three-cycles are conjugate to their permutations.
• $(abc)$ is conjugate to $(abd)$ by $(cde)$; this covers the case of introducing a single new element to the cycle.
• $(abc)$ is conjugate to $(ade)$ by $(bd)(ce)$; this covers the case of introducing two new elements to the cycle.

Five-cycles

This class does split: I claim that $(12345)$ and $(21345)$ are not conjugate. (Once we have this, then the class must split into two chunks, since $\{ \rho (12345) \rho^{-1}: \rho \ \text{even} \}$ is closed under conjugation in $A_5$, and $\{ \rho (12345) \rho^{-1}: \rho \ \text{odd} \}$ is closed under conjugation in $A_5$. The first is the conjugacy class of $(12345)$ in $A_5$; the second is the conjugacy class of $(21345) = (12)(12345)(12)^{-1}$. The only question here was whether they were separate conjugacy classes or whether their union was the conjugacy class.)

Recall that $\tau (12345) \tau^{-1} = (\tau(1), \tau(2), \tau(3), \tau(4), \tau(5))$, so we would need a permutation $\tau$ such that $\tau$ sends $1$ to $2$, $2$ to $1$, $3$ to $3$, $4$ to $4$, and $5$ to $5$. The only such permutation is $(12)$, the transposition, but that is not actually a member of $A_5$.

Hence in fact $(12345)$ and $(21345)$ are not conjugate.