Q:

Find the conjugacy class $O_{(234)}$ if $G=A_{4}$ is acting on itself by conjugation.

Find the conjugacy class $O_{(234)}$ if $G=A_{4}$ is acting on itself by conjugation.

I calculated $(x)(234)$ for all $x\in A_{4}$ and got the set $O_{(234)}=\left \{(234), (143), (142), (123), (132)\right \}$. 
This is because $(1)(234) = (234)$, $(123)(234) = (143)$, $(132)(234) = (241) = (142)$, $(142)(234) = (321) = (123)$, and $(143)(234) = (132)$.
Am I missing anything or doing anything wrong?

A:

Given that $|A_4|=12$ it isn't too much work to explicitly write out all conjugations. A few checks for your work:

As noted in the comments, the size of the conjugacy class needs to divide $|A_4|=12$.
Cycle type is preserved by conjugation, so all elements of the conjugacy class are $3$-cycles.

A bit more thought gets you even further; as there are precisely eight $3$-cycles in $A_4$, there are at most eight conjugates. As the conjugacy classes partition the $3$-cycles and $S_4$ acts transitively on them (because it acts transitively on $3$-cycles), the size of the conjugacy classes must also divide $8$. So the number of conjugates divides $\gcd(8,12)=4$.
Alternatively, by the orbit-stabilizer theorem, as the subgroup generated by $(2\ 3\ 4)$ clearly stabilizes $(2\ 3\ 4)$, and this subgroup contains $3$ elements, there are at most $\frac{|A_4|}{|\langle(2\ 3\ 4)\rangle|}=\frac{12}{3}=4$ conjugates.

