Finding the centraliser of an element in $S_5$ 
Let $x = (12)(34) \in S_5$. Is there a quick way to find $C_{S_5}(x)$ ? 

I know that conjugation leaves cycle types unchanged. Thus, if $\sigma \in C_{S_5}(x)$, then $(12)(34) = \sigma^{-1}(12)(34)\sigma = (\sigma(1), \sigma(2))(\sigma(3), \sigma(4))$, but how does this help me?
Any help would be appreciated. Thanks!
 A: There may be a faster way to do this, but the first that comes to mind is to use the fact that $(\sigma(1)\sigma(2))(\sigma(3),\sigma(4)) = (12)(34)$ to determine which $\sigma$ are possible. 
Since $\sigma$ is a bijection, $(\sigma(1),\sigma(2))(\sigma(3),\sigma(4))$ is written in disjoint cycle notation. Thus $(\sigma(1),\sigma(2)) = (1,2)$ or $(3,4)$ and $(\sigma(3),\sigma(4)) = (3,4)$ or $(1,2)$. In any case we see $\sigma$ must fix $5$. We can choose any value for $\sigma(1)$ in $\{1,2,3,4\}$. This then determines the value of $\sigma(2)$. Once we've chosen $\sigma(1)$. The choice of $\sigma(3)$ determines $\sigma(4)$. With these observations we can list the elements. 
$C_{S_5}((12)(34)) = \{1, (34), (12), (12)(34), (13)(24), (1324), (1423), (14)(23)\}$
A: For $\sigma \in S_5$ we have that $(12)(34) = \sigma^{-1}(12)(34)\sigma = (\sigma(1), \sigma(2))(\sigma(3), \sigma(4))$ holds if and ony if $\{\sigma(\{1,2\}),\sigma(\{3,4\})\}=\{\{1,2\},\{3,4\}\}$.
Thus we have two possibilities: either $\sigma(\{1,2\})=\{1,2\}$ and $\sigma(\{3,4\})=\{3,4\}$ or $\sigma(\{1,2\})=\{3,4\}$ and $\sigma(\{3,4\})=\{1,2\}$.
In the first case, we get the possibilities $\mathrm{id},(12),(34),(12)(34)$, in the second case, we get the possibilities $(13)(24),(14)(23),(1324),(1423)$
