= Solution
Every finite <cyclic subgroup> of order $m$ has exactly $\varphi(m)$ generators, where $\varphi$ 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 $\varphi(m)$.
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 is
$$
\begin{array}{c|r|r|r}
\text{cycle type}&\text{elements}&\text{order}&\text{cyclic subgroups}\\\hline
1^5&1&1&1\\
2\,1^3&10&2&10\\
2^2\,1&15&2&15\\
3\,1^2&20&3&10\\
3\,2&20&6&10\\
4\,1&30&4&15\\
5&24&5&6
\end{array}
\qquad \boxed{1+10+15+10+10+15+6=67.}
$$
The two <subgroups> $\langle(12)\rangle$ and $\langle(12)(34)\rangle$ 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>.
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