 Solution

Every finite cyclic subgroup of order $m$ has exactly $\phi (m)$ generators, where $\phi$ is the Euler totient function: if $g$ has order $m$, then $g^k$ generates precisely when $\gcd (k,m)=1$. Each element generates exactly one cyclic subgroup. Thus counting cyclic subgroups by their generators amounts to counting elements by order of a group element and dividing by $\phi (m)$.

Use the cycle decomposition of a permutation in $S_5$. A permutation with $a_j$ cycles of length $j$ occurs $5!/{\prod }_j(j^{a_j}{a}_j!)$ times: order the five letters, then divide by the cyclic choices of starting point and the orderings of equal-length cycles. Its order is the least common multiple of the cycle lengths. The resulting count isThe two subgroups $⟨(12)⟩$ and $⟨(12)(34)⟩$ are both isomorphic to $C_2$ by a group isomorphism. They are not conjugate subgroups: conjugating their unique nonidentity elements would have to preserve cycle type, whereas a transposition and a product of two disjoint transpositions have different cycle types.