Why is there no Latin square of order 7 with an autotopism group of order 7? 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 $11$.
 A: I can show that it is enough to look through all automorphisms (according to your definition) and thus it is enough to look through 7! examples for 7:
Suppose that we have a Latin square of order $k$ with an autotopism of order $k$.
We cannot have two of the three permutations be the identity (because only changing one of the three things clearly changes the Latin square).
If just the last permutation is the identity, then we can without loss of generality assume that the other two permutations are cycles $(1,2,\dots,k)$ (this is just a conjugation of the isotopy group). This shows that the letters in the square are constant on the extended diagonals and this immediately gives an isotopy by cyclic permutation of the rows and then cyclic permutation of the letters according to their order in the first column.
So, we found an additional isotopy that is not in the cyclic group generated by the original one.
If one of the two other permutations is the identity, the argument is analogous because the letters can be interpreted as height in a cube.
So, we are left with triples of $k$-cycles. Without loss of generality, it is enough to regard all cycles equal to $(1,2,\dots,k)$ which shows that each diagonal has to read $1,2,\dots,k$. There are $k!$ of them if you fix the positions of the 1s as an arbitrary permutation.
It might be possible to show that these objects have additional symmetries for certain values of $k$.
