Let be $G$ and $H$ two equinumerous groups of order $n$. We label the elements in such a way that $$ 1< |g_1|\le|g_2|\le \cdots \le |g_n| $$ $$ 1< |h_1|\le|h_2|\le \cdots \le |h_n| $$ If $(|g_1|,|g_2|,\cdots,|g_n|)\ne (|h_1|,|h_2|,\cdots,|h_n|)$ then $G\not\cong H$.
Is there a result deducing from $(|g_1|,|g_2|,\cdots,|g_n|)= (|h_1|,|h_2|,\cdots,|h_n|)$ something about the isomorphism of $G$ and $H$?
It does not follow that $G$ and $H$ are isomorphic from the given hypothesis. As indicated in this answer to a similar question (https://math.stackexchange.com/a/1478993/81163), $G = \mathbb{Z}/4\times \mathbb{Z}/4$ and $H = \mathbb{Z}/2 \times Q_8$. where $Q_8$ is the quaternions, are not isomorphic, but they have the same number of elements of each order.