The group $S_n \times S_n \times S_n$ acts on the set of Latin squares $L$ of order $n$, with $\theta:=(\alpha,\beta,\gamma)$, starting from $L$, permuting the rows by $\alpha$, permuting the columns by $\beta$ and permuting the symbols by $\gamma$. Each $\theta \in S_n \times S_n \times S_n$ is called an isotopism. In some instances $\theta(L)=L$, whence we call $\theta$ an autotopism of $L$. The set of autotopisms of a Latin square forms a group under composition.
[Side note: we reserve the name automorphism for autotopisms of the form $(\alpha,\alpha,\alpha)$ to correspond with the notions in group theory and quasigroup theory]
From Brendan McKay's data, we can deduce that no Latin square of order 7 has an autotopism group of order 7. [There are autotopisms of order 7, but no autotopism group has order 7.]
What's a clever proof of this observation that doesn't look through all the non-isotopic Latin squares?
Alexander Hulpke, Petteri Kaski, Patric R. J. Östergård, The number of Latin squares of order 11, claims that there is a Latin square of order 11 that admits an autotopism group of order $11$.