Claim: If there is more than one hedgehog, it is impossible for them all to get distinct numbers. Proof: If no two hedgehogs share a row or column, then clearly every hedgehog gets the number $1\cdot1 = 1$, which is no good. Thus we may assume there is some row or column with at least $2$ hedgehogs. Consider a row or column (assume wlog it is a row) containing the maximal number of hedgehogs, say, $n$ hedgehogs where $n \geq 2$. The $n$ columns containing those hedgehogs must all have distinct numbers of hedgehogs, but they each have at most $n$ hedgehogs. Hence, the columns have $1, 2, \dots, n$ hedgehogs in some order. In particular, the hedgehog in the $1$-hedgehog column gets the number $n\cdot1 = n$. Applying the same logic to the column with $n$ hedgehogs, we find that the $n$ rows containing these hedgehogs have $1, 2, \dots, n$ hedgehogs in some order, so the hedgehog in the $1$-hedgehog row gets the number $1\cdot n = n$. Thus, there are two distinct hedgehogs that both get the number $n$. (Note that they are necessarily distinct because $n \neq 1$.)