If $sigma in A_n$ has a split conjugacy class, then $tau sigma tau^{-1}$ is not conjugate to $sigma$ for odd $tau$

Let $sigma in A_n$ such that the disjoint cycle decomposition of $sigma$ contains no even cycles or cycles of the same length. Although there is one conjugacy class of $sigma$ in $S_n$, there are two conjugacy classes of $sigma$ in $A_n$. We can choose any permutation in the original $S_n$ conjugacy class to represent one of the two $A_n$ classes. But how can we find another permutation to represent the other $A_n$ class?
These answers suggest the following to be true:

Let $sigma in A_n$. If the conjugacy class of $sigma$ splits in $A_n$, then $tau sigma tau^{-1}$ is not conjugate (in $A_n$) to $sigma$ for all $tau in S_n setminus A_n$.

What is a proof of this fact?

From here we know that $tau sigma tau^{-1} neq sigma$ for all odd $tau$. But proving $tau sigma tau^{-1} $ is not conjugate (in $A_n$) to $sigma$ seems to be a stronger claim.