I'm having a really hard time understanding the way $A_n$ the alternating group behaves. I understand it means that $\forall \sigma \in A_n$ we have $sgn(\sigma) = 1$ but I'm not sure I understand the way the sgn function works.
Anyway the hint in the question is to look for some $\lambda \in S_5$ that satisfies $\lambda \sigma \lambda^{-1} = \tau \iff \lambda \notin A_5$
So first of all I know I need to find $\tau \in S_5$ such that $|\tau| = 5$ meaning it's cycle is the length of 5 because that is the only way to have $\tau$ conjugated to $\sigma$ but how can I make sure $\tau \notin A_5$ ?
First, we use Orbit-Stabilizer theorem in $S_5$,
$$|C_{S_5}(12345)|=\frac{|S_5|}{|cl_{S_5}(12345)|}=\frac{5!}{|cl_{S_5}(12345)|}$$
where $|cl_{S_5}(12345)|$ is the number of elements in the conjugacy class of $(12345)$ in $S_5$, we know:
$$|cl_{S_5}(12345)|=\frac{5!}{5^1\cdot 1!}=4!$$
Plug in and we get:
$$|C_{S_5}(12345)|=5$$
Next, we move to $A_5$. Since $cl_{S_5}(12345)$ splits in $A_5$, we get
$$|cl_{A_5}(12345)|=\frac{1}2|cl_{S_5}(12345)|=12$$
Again, we use Orbit-Stabilizer Theorem,
$$|C_{A_5}(12345)|=\frac{|A_5|}{|cl_{A_5}(12345)|}=\frac{5!/2}{12}=5$$
So all centralizers belong to $A_5$, and the answer is zero.
Remarks:
Actually, all the five centralizers form the cyclic group $\langle(12345)\rangle$, and they are all even permutations, hence all belong to $A_5$.