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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!

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 $(13)(24),(14)(23),(1324),(1423)$

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.

$C_{S_5}((12)(34)) = {1, (34), (12), (12)(34), (13)(24), (1324), (1423), (14)(23)}$