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Jun 4 at 12:16 comment added Igor Khavkine The structure of both the automorphisms and their orbits is well-known in the literature. I've added some references at the end of my answer.
Jun 4 at 10:24 comment added Emil Jeřábek I’ll stop the edits. But the criterion generalizes as follows: tuples $\{x_i:i<t\},\{y_i:i<t\}\in G^t$ are automorphic iff for each prime power $p^l$ that divides $|G|$, the assignment $x_i+p^lG\mapsto y_i+p^lG$, $i<t$, extends to an isomorphism $(\langle\vec x\rangle+p^lG)/p^lG\simeq(\langle\vec y\rangle+p^lG)/p^lG$.
Jun 4 at 9:50 history edited Emil Jeřábek CC BY-SA 4.0
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Jun 4 at 9:23 comment added Emil Jeřábek Yes. I’m just not entirely sure this is trivial for everyone.
Jun 4 at 9:20 history edited Emil Jeřábek CC BY-SA 4.0
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Jun 4 at 9:20 comment added YCor Well, this is just a trivial fact, isn't it? Just write a 1st order sentence saying, for the given group of order $n$, "there are $n$ distinct elements, and no other one, satisfying the given group law".
Jun 4 at 8:10 comment added Emil Jeřábek I realized the second link is quite useless, as it only shows the part of the argument that's irrelevant for structures in a finite language such as groups. I'll see if I can find a better one.
Jun 4 at 8:04 comment added Emil Jeřábek If you want to express it in terms of order in $G/p^lG$, you don't need $k$: it just says that $x$ and $y$ have the same order in $G/p^lG$.
Jun 4 at 8:00 comment added YCor Note: (2) can be restated as: for each $k,\ell$, $p^kx$ and $p^ky$ have the same order in $G/p^\ell G$.
Jun 4 at 7:16 history edited Emil Jeřábek CC BY-SA 4.0
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Jun 4 at 7:10 history answered Emil Jeřábek CC BY-SA 4.0