Where C is the centralizer in $S_5$.
Obviously, $\forall i \in \mathbb{Z}, \ (12345)^i(12345)(12345)^{-i} = (12345) \Longrightarrow \langle(12345)\rangle \subset C_{S_5}((1,2,3,4,5))$.
The possible cycle structures in $S_5$ are (1,1,1,2), (1,1,3), (1,4), (5), (1,2,2) and (2,3). However, it's not only the structure of $\sigma \in S_5$ that might be relevent to whether or not $\sigma \in C_{S_5}((12345))$, but also the exact permutation (e.g. maybe (123) will affect (12345) differently from (132)), as far as I can see.
Is there any elegant way to prove the equalty, rather then checking for each $\sigma \in S_5$?
Suppose that $h^{-1} g h = g$, where $g=(12345)$, then $(h(1)h(2)h(3)h(4)h(5)) = (12345)$. Now, we can choose $h(1)$ in any of $5$ ways, but once these are chosen, the rest are fixed, because $h(2) = h(1) + 1$ and so on so forth. Hence, there are precisely $5$ elements $h$ that commute with $g$. However, we have already figured out five such elements, namely the powers of $(12345)$. Hence, it follows that the centrlizer consists precisely of these elements.