Step 1: Let $G$ be a group with $|G|=30=2\cdot3\cdot5$ and let $P$ be a Sylow 5-subgroup, so $|P|=5$. By definition, the number of Sylow 5-subgroups is $n_5=[G:N_G(P)]$, the index of the normalizer of $P$ in $G$.
Step 2: Since $P\subseteq N_G(P)\subseteq G$, the order $|N_G(P)|$ is a multiple of $|P|=5$ and a divisor of $|G|=30$. So $|N_G(P)|\in\{5,10,15,30\}$, giving $n_5=30/|N_G(P)|\in\{6,3,2,1\}$.
Step 3: Sylow's theorem also requires $n_5\equiv 1\pmod5$. Testing the four candidates: $6\equiv1$, $3\equiv3$, $2\equiv2$, $1\equiv1\pmod5$.
Step 4: Only $n_5=6$ and $n_5=1$ satisfy the congruence, so these are the only possible values.
\[\boxed{n_5 = 1 \text{ or } 6}\]