Let $\sigma\in S_n$ be an even permutation. The size of its conjugcay class in $S_n$ is, by the orbit-stabilizer theorem, equal to $$|[\sigma]_{S_n}|=|S_n|/|C_{S_n}(\sigma)|.\qquad(*)$$
Similarly, in $A_n$ we get
$$
|[\sigma]_{A_n}|=|A_n|/|C_{A_n}(\sigma)|.\qquad(**)
$$
Here $|A_n|=|S_n|/2$ and $C_{A_n}(\sigma)=C_{S_n}(\sigma)\cap A_n$. In other words, $C_{A_n}(\sigma)$ consists of the even permutations in $C_{S_n}(\sigma)$. It follows that either $C_{A_n}(\sigma)=C_{S_n}(\sigma)$, or $C_{A_n}(\sigma)$ is an index two subgroup of $C_{S_n}(\sigma)$. The conjugacy class splits in two in the former case (see equations $(*)$ and $(**)$). We also see that the latter case (=no splitting) occurs, when there are odd permutations in $C_{S_n}(\sigma)$.
The conjugacy classes of an even permutation in $A_n$ is equal to its conjugacy class in $S_n$ if and only if it is centralized by at least one odd permutation. OTOH, if the centralizer only contains even permutations, then the conjugacy in $S_n$ splits into two equal size conjugacy classes of $A_n$.
The 2-cycle $(9;10)$ is in the centralizer $C_{S_{11}}((12345)(678))$. This means that the conjugacy class of $\sigma=(12345)(678)$ does not split in $A_{11}$.
