= Solution
The <normalizer> is $N_G(H)=\{g\in G:gHg^{-1}=H\}$. It is a <subgroup>: the identity normalizes $H$, products of normalizing elements normalize it, and conjugating the equality by $g^{-1}$ proves inverse closure. The map
$$
G/N_G(H)\longrightarrow\{\text{conjugates of }H\},\qquad gN_G(H)\longmapsto gHg^{-1}
$$
is well-defined and onto. Two images are equal exactly when $g_2^{-1}g_1\in N_G(H)$, exactly the condition that their <cosets> coincide. Thus \b[the number of conjugates is $\boxed{[G:N_G(H)]}$].
By the permitted conjugacy assertion for <Sylow subgroups>, the number $n_p$ of Sylow $p$-<subgroups> equals $[G:N_G(P)]$ for one such <subgroup> $P$. <Lagrange's theorem> makes this divide $|G|$, and $P\subseteq N_G(P)$ also makes it divide $|G|/|P|$.
For order $72=8\cdot9$, a Sylow $3$-<subgroup> has order nine, so $n_3$ divides eight. To obtain the required congruence without assuming an additional Sylow counting theorem, let $P$ act by conjugation on the set of Sylow $3$-<subgroups>. Its orbit sizes are powers of three. A fixed <subgroup> $Q$ is normalized by $P$, so $PQ$ is a <subgroup> and its order $|P||Q|/|P\cap Q|$ is a power of three. Since nine is the largest power of three dividing $72$, this forces $PQ=P=Q$. There is exactly one fixed point, namely $P$ itself. All other orbits have sizes divisible by three, giving $n_3\equiv1\pmod3$. Among $1,2,4,8$, only one and four satisfy this. Therefore \b[$\boxed{n_3\in\{1,4\}}$].
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