Centralizer and Normalizer of permutation in $S_{12}$

Question

Let $\sigma = (1\,2\,\ldots\,6)(7 \,8 \ldots\,12) \in S_{12}$, a permutation of two distinct 6-cycles in $S_{12}$.

1. Find the size of the centralizer of $\sigma$, $|C_{S_{12}}(\sigma)|$. Prove it.
2. Find the size of the normalizer $N_{S_{12}}(\langle\sigma\rangle)$. Prove it.

I tried to show as follows, not sure about the proof in (2):

1. G acts on G with conjugation. $|C_{S_{12}}|=\frac{|G|}{|x^G|}$. There are $\frac{1}{2}\cdot\binom{12}{6}\cdot5!\cdot5!=\frac{12!}{72}$ conjugations, as conjugation has same cycles structure. Therefore, $|C_{S_{12}}|=72$.
2. $N_{S_{12}}(\langle\sigma\rangle) = \{{g \in G \mid g \langle \sigma \rangle}=\langle \sigma \rangle g \}$. Then $\operatorname{ord}(g \sigma g^{-1}) = \operatorname{ord}(\sigma ^ i)$, then $\gcd(i,6)=1$, so $i \in \{1,5\}$. For $i=1$, that's all of $C_{S_{12}}(\sigma)$. Otherwise, we can show every $g,g'$ such that $g \sigma g^{-1} = g' \sigma {g'}^{-1} = \sigma^5$ creates the same left coset $g C_{S_{12}}(\sigma)$. So there are $72 \cdot 2$ permutations in the normalizer, 144 in total.