$σ^2 = (a_1, a_3, a_5,\dots , a_{2k+1}, a_2, a_4, \ldots , a_{2k})$.
I'm having a bit of trouble understanding this question. If I were to square a cycle with an odd order and end up with this result, wouldn't this mean that the order of $σ^2$ is even and therefore, not the same?
Order of $\sigma$ is $m$. Let the order of $\sigma^2$ be $k$. Note that $k \leq m$ as $(\sigma^2)^m=(\sigma^m)^2=e$.
Now, $(\sigma^2)^k=e$ implies that $m$ divides $2k$. But $m$ is odd, and hence this implies that $m$ divides $k$. But $k \leq m$ which implies that $m=k$.
Without loss of generality, let $σ = (123 · · · m)$ where $m$ is odd. Then, $σ^2 = (123 · · · m)(123 · · · m) = (135 · · · m246 · · ·(m − 1))$. QED
Given , $\color{blue}{\sigma \in S_n}$ and $\color{blue}{o(\sigma) =m}$
Then, $\color{red}{\langle \color{blue}\sigma\rangle}$ is a cyclic group of order $m$.
Then for any $k\in \Bbb{Z}$ , $$\color{red}{\boxed{\color{blue}{o(\sigma^k) =\frac{m}{\color{red}{\gcd(k,m}}}}}$$
For the proof of the above theorem see here :
How to prove $|a^k|=n/\gcd(n,k)$ whenever $|a|=n$?
Now, $\color{blue}{o(\sigma^2)=\frac{m}{\color{red}{\gcd(2, m) }}}$
Given $m$ is odd. Hence, $\color{red}{\gcd(2, m) =1}$
Hence, $\color{blue}{o(\sigma^2) =m}$
