For what values on $n$ can we find an order $n$ Latin square that does not come from a group table? I understand that most Latin squares can be constructed from group tables...
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For all orders $n \geq 5$.
Proof: Latin squares that are isotopic to group tables satisfy what's known as the quadrangle criterion. Let L be a Latin square in which, for any pair of 2x2 submatrices that have the form
we have x=d. Then we say L satisfies the quadrangle criterion. In particular, Latin squares that are isotopic to group tables satisfy the quadrangle criterion. (See e.g. the Denes and Keedwell book, among other references.)
So, to construct a Latin square that's not isotopic to a group table, we just break the quadrangle criterion.
Start with the first two rows
It does not satisfy the quadrangle criterion since it contains the submatrices
We can complete the first two rows to a $2 \times n$ Latin rectangle provided $n \geq 7$ (or $n=5$). Then by Hall's Theorem, this completes to a Latin square. Since we've broken the quadrangle criterion, it's not isotopic to a group table. (I'll let you find an example of order 6.)
In response to the claim in the question, the complete opposite is true, and can be proved via the following result:
Theorem (McKay, Wanless 2005): Almost all Latin squares have a trivial autotopism group.
Since groups of orders $n \geq 3$ have non-trivial automorphism groups, they thus have non-trivial autotopism groups, and thus, almost all Latin squares are not isotopic to group tables.