Let $G=\langle\mu\rangle$ be the subgroup of $S_4$, now compute the coset of $G$ using...? 
Let $G= \langle \mu \rangle$ be the subgroup $S_4$ generated by $$\mu =\begin{pmatrix}
1 & 2 & 3 & 4\\
2 & 3 & 4 & 1
\end{pmatrix}.$$
Compute all the cosets of G containing the permutation $\begin{pmatrix}
1 & 2 & 3 & 4\\
4 & 3 & 2 & 1
\end{pmatrix}.$

$$G=  \left\{\begin{pmatrix} 1 & 2 & 3 & 4 \\
1 & 2 & 3 & 4\end{pmatrix}, \begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 3 & 4 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 1 & 2 \end{pmatrix}, \begin{pmatrix} 1 & 2 & 3 & 4 \\ 4 & 1 & 2 & 3 \end{pmatrix}\right\}$$
Coset of G is simply $ga$ $\forall g \in G$
Therefore we would have

*

*$\begin{pmatrix} 1 & 2 & 3 & 4 \\
1 & 2 & 3 & 4\end{pmatrix} \begin{pmatrix}
1 & 2 & 3 & 4\\
4 & 3 & 2 & 1
\end{pmatrix} = \begin{pmatrix}
1 & 2 & 3 & 4\\
4 & 3 & 2 & 1
\end{pmatrix}$


*$\begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 3 & 4 & 1 \end{pmatrix}\begin{pmatrix}
1 & 2 & 3 & 4\\
4 & 3 & 2 & 1
\end{pmatrix}=(1 3)(2 3)(4 1)  (4 1)(2 3)=\begin{pmatrix} 1 & 2 & 3 & 4 \\
3 & 2 & 1 & 4\end{pmatrix}$


*$\begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 1 & 2 \end{pmatrix}\begin{pmatrix}
1 & 2 & 3 & 4\\
4 & 3 & 2 & 1
\end{pmatrix}=(1 3) (4 2)  (4 1)(2 3)=\begin{pmatrix} 1 & 2 & 3 & 4 \\
2 & 1 & 3 & 4\end{pmatrix}$


*$\begin{pmatrix} 1 & 2 & 3 & 4 \\ 4 & 1 & 2 & 3 \end{pmatrix} \begin{pmatrix}
1 & 2 & 3 & 4\\
4 & 3 & 2 & 1
\end{pmatrix}= (2 3) (4 3)(4 1)  (4 1)(2 3)=\begin{pmatrix} 1 & 2 & 3 & 4 \\
1 & 4 & 3 & 2\end{pmatrix}$
So I am not $100\%$ sure if I am thinking about this correctly. Please let me know if there are other ways or more clear ways of thinking about this.
I think if $\mu$ as a $90$ degree rotation counter clockwise to a square that has
one of the left top corner, two in the top right corner, 3 on the bottom right corner and 4 on the bottom left corner. Though I don't think this is the right way of thinking about this, especially when applying $a$ to $ga$ (because $a$ would be a $90$ degree rotation clockwise)?

Hint: Cosets of any group by any subgroup partition the group . . .

 . . . and so each element of the desired cosets of $G$ is in exactly one coset; namely, $G\begin{pmatrix} 1 & 2 & 3 & 4\\ 4 & 3 & 2 & 1 \end{pmatrix}$.

