Given a permutation $\sigma = (13)(254)$, state $\sigma^2$. Given a permutation $\sigma = (13)(254)$, state $\sigma^2$.
$\sigma = (13)(254), \sigma^2=(13)(254)(13)(254) = (13)(13)(254)(254) = (425)
$
Or, in two row format, get:
$$ \sigma = \begin{pmatrix} 1 & 3 \\ 3 & 1 \end{pmatrix}\begin{pmatrix} 2 & 5 &4 \\ 5 & 4 & 2\end{pmatrix} $$
Then, $$\sigma^2=\begin{pmatrix} 1 & 3 \\ 3 & 1 \end{pmatrix}\begin{pmatrix} 2 & 5 &4 \\ 5 & 4 & 2\end{pmatrix}
\begin{pmatrix} 1 & 3 \\ 3 & 1 \end{pmatrix}\begin{pmatrix} 2 & 5 &4 \\ 5 & 4 & 2\end{pmatrix}
$$
Due to disjoint cycles, they can commute as no affect on map produced.
$$\sigma^2=\begin{pmatrix} 1 & 3 \\ 3 & 1 \end{pmatrix}\begin{pmatrix} 1 & 3 \\ 3 & 1 \end{pmatrix}
\begin{pmatrix} 2 & 5 &4 \\ 5 & 4 & 2\end{pmatrix}
 \begin{pmatrix} 2 & 5 &4 \\ 5 & 4 & 2\end{pmatrix}
$$
My answer is:
$$\sigma^2=\begin{pmatrix} 1 & 3 \\ 1 & 3 \end{pmatrix}
\begin{pmatrix} 2 & 5 &4 \\ 4 & 2 & 5\end{pmatrix}
= e.\begin{pmatrix} 2 & 5 &4 \\ 4 & 2 & 5\end{pmatrix}
= \begin{pmatrix} 2 & 5 &4 \\ 4 & 2 & 5\end{pmatrix}
$$
But, the answer is different: $(245)$?

Here
$$\begin{align}
\sigma^2&=\begin{pmatrix}1&2&3&4&5\\3&5&1&2&4\end{pmatrix}\cdot\begin{pmatrix}1&2&3&4&5\\3&5&1&2&4\end{pmatrix}\\
&=\begin{pmatrix}1&2&3&4&5\\1&4&3&5&2\end{pmatrix}\\
&=(245).
\end{align}$$
