How many elements of $S_9$ commute with $(123)(4567)$?

Consider the action of $ S_9 $ on itself, given by conjugation. The stabilizer of an element $ x \in S_9 $ is its centralizer, and the orbit-stabilizer theorem gives

$$ 9! = |S_9| = |C(x)| \cdot |S| $$

where $ S $ denotes the set of conjugates of $ x $, and $ C(x) $ is the centralizer. Conjugate elements in $ S_n $ are precisely those which have the same cycle type, therefore, $ S $ is the set of all $ 3, 4 $ cycles when $ x = (123)(4567) $. By elementary combinatorics, there are exactly

$$ \binom{9}{3} \cdot 2! \cdot \binom{6}{4} \cdot 3! = 15120 $$

such cycles. Thus,

$$ C((123)(4567)) = \frac{9!}{15120} = 24 $$

elements of $ S_9 $ commute with $ (123)(4567) $.