 Solution

Compose permutations from right to left. Tracking the images gives $1\mapsto 2\mapsto 1$, $3\mapsto 4\mapsto 3$, and $5\mapsto 5$, soTwo transpositions give sign of a permutation $+1$, proving that this element belongs to the alternating group $A_5$.

Its conjugacy class consists of all double transpositions. Conjugation merely relabels the letters, so no other cycle type can occur. Conversely, choose a permutation $h\in S_5$ relabelling the two pairs into any desired two pairs. If $h$ is odd, replace it by $h(12)$: the transposition $(12)$ centralizes $(12)(34)$, so this replacement is even and gives the same conjugate. Thus every double transposition is reached by conjugation inside $A_5$.

The complete list of fifteen elements, grouped by their fixed letter, isThere are five choices of fixed letter and three pairings of the other four letters, confirming the conjugacy class size $15$.