<s>[INST] Find permutation that solves $\;\tau \circ X = \sigma$ I need to find a permutation $X$ that solves $\;\tau \circ X = \sigma,\;$ given 
$$\tau = \begin{bmatrix} 
1 & 2 & 3 & 4 & 5\\
3 & 4 & 5 & 2 & 1
 \end{bmatrix} = (1,3,5)(2,4)$$
$$\sigma = \begin{bmatrix} 
1 & 2 & 3 & 4 & 5\\
3 & 2 & 1 & 5 & 4
 \end{bmatrix} = (1,3)(2)(4,5)$$
Is there a trick on how to solve for $X$?
 [/INST] Assuming you compose left to right:
$$\tau = \begin{bmatrix} 
1 & 2 & 3 & 4 & 5\\
3 & 4 & 5 & 2 & 1
 \end{bmatrix} = (1,3,5)(2,4)$$
times
$$X = \begin{bmatrix} 
3 & 4 & 5 & 2 & 1\\
3 & 2 & 1 & 5 & 4
 \end{bmatrix} = (1,4,2,5)(3)$$
equals
$$\sigma = \begin{bmatrix} 
1 & 2 & 3 & 4 & 5\\
3 & 2 & 1 & 5 & 4
 \end{bmatrix} = (1,3)(2)(4,5)$$
</s>