Suppose $H$ and $G$ are finite groups. Define $s(G)$ as the number of subgroups of $G$. Is it true, that if $|H| = |G|$ and $s(G) = s(H)$ then $G \cong H$?
This conjecture holds for $|G| = |H| \leq 10$:
For $|G|=|H| \in \{1, 2, 3, 5, 7\}$ there is only one group of the corresponding order.
For $|G| = |H| = 4$:
$$s(C_4) = 3$$
$$s(C_2 \times C_2) = 5$$
For $|G| = |H| = 6$:
$$s(C_6) = 4$$
$$s(S_3) = 6$$
For $|G| = |H| = 8$:
$$s(C_8) = 4$$
$$s(C_2 \times C_4) = 8$$
$$s(C_2 \times C_2 \times C_2) = 16$$
$$s(D_8) = 10$$
$$s(Q_8) = 6$$
For $|G| = |H| = 9$:
$$s(C_9) = 3$$
$$s(C_3 \times C_3) = 5$$
For $|G| = |H| = 6$:
$$s(C_6) = 4$$
$$s(D_5) = 8$$
However, I do not know, whether this pattern continues to hold, or there is a relatively large pair of counterexamples.