Question about the conjugation of an element in a group Let $(a_1,a_2,a_3)$ be a 3-cycle in the alternating group $A_4$ in four letters. Find $g \in A_4$ such that 
$$g(a_1,a_2,a_3)g^{-1}=(a_2,a_1,a_4) = (a_1,a_2,a_4)^{-1}$$
Why do we need the last equality? What I am doing doing is since $g(a_1,a_2,a_3)g^{-1} = (g(a_1),g(a_2),g(a_3))$. I can set $g = (a_1,a_2),(a_3,a_4)$ and this will give me $(a_2,a_1,a_4)$. However, if I do it on the last equality instead, since $(a_1,a_2,a_4)^{-1} = (a_4,a_2,a_1)$, I will get $g=(a_1,a_4,a_3)$. I am wondering why is this contradiction?
Or is it not contradiction? Just there are two different answers?
 A: You want to find an element $g \in A_4$ such that $g(123)g^{-1}=(214)$.  It is true that more than one solution is possible. For example, take $g=(12)(34)$ or $g=(143)$, as you showed.  Any two solutions $g$ and $h$ would satisfy the equality $gag^{-1}=hah^{-1}$.  This implies $h^{-1}g a = a h^{-1}g$, i.e. $h^{-1}g$ commutes with $a$.  In your case $h^{-1}g = ((12)(34))^{-1} (143)$ is a power of $(123)$ and hence commutes with $(123)$.  
The set of elements in $A_4$ that commute with $a$ is called the centralizer of $A_4$ in $a$, denoted $C_{A_4}(a) =: C$.  Conjugating $a$ by $g$ gives the same value as conjugating $a$ by $h$ iff $h^{-1}g \in C$, iff $g \in Ch$, i.e. $g$ and $h$ must lie in the same coset of $C$ in $A_4$.  Since $a=(123)$ in this example, the centralizer in $A_4$ of $a$ is the cyclic group $\langle (123) \rangle$ containing three elements.  After you found one of the solutions $h=(12)(34)$, all other solutions can be found by computing $hC:=\{hc: c \in C\} = \{(12)(34), (12)(34)(123), (12)(34)(132)\} = \{(12)(34), (243), (143)\}$.  
