Let $\sigma = (1 2 \dots 9) \in S_{10}$.
a) Calculate the size of the normalizer $N_{S_{10}}(<\sigma >)$.
b) Describe exactly the elements in $N_{S_{10}}(<\sigma >)$.
I am not sure how to approach this. I understand that we look for permutations that fixes $10$, and yet I can't see what to do further...
I know that $\tau \sigma \tau^{-1} =(\tau(1) \dots \tau(9))$ by definition, and also that for $\tau$ to be in $N_{S_{10}}(<\sigma >)$ than it is required that $\tau \sigma \tau^{-1} =(\tau(1) \dots \tau(9)) = \sigma^i$ for some $i$.
How can I continue from here?
I'll try to continue from where you stopped...
It has to be $o(g\sigma g^{-1})=o(\sigma^i)=9$ so $gcd(i,9)=1\Rightarrow i=1,2,4,5,7,8$
The $g\in S_{10}$ with the above property (for fixed $i$) form a coset of $C_{S_{10}}(<\sigma>)=<\sigma>$ and the number of such cosets is $6$ (because we have $6$ $i$'s)
For example for the $g\in S_{10}$ with $g\sigma g^{-1}=\sigma^2$ we see that all $t\in S_{10}:t\sigma t^{-1}=\sigma^2$ are $gC(<\sigma>)=\{t: t=gu \text{ with } u\sigma u^{-1}=\sigma\}$
And since $|<\sigma >|=9$, $|N_{S_{10}}(<\sigma>)|=9\cdot 6=54$
For (b) the elements of $N_{S_{10}}(<\sigma>)$ are these of the form $g\sigma^l$ with $\sigma^l\in<\sigma> $ and $g\in S_{10}$ with the property described in the first line.
I hope it is correct!