2
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I know that automorphism group of $\mathbb Z_q$ is $(q-1)$ because the group is cyclic and its generator's image determines the automorphism and we have exactly $(q-1)$ choices.

For $\mathbb Z_q \times\mathbb Z_q$ I was thinking that since the group is generated by $\{(0,1),(1,0)\}$ but couldn't think how to see the maps that would be automorphisms.

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    $\begingroup$ Any such map is an automorphism of $\mathbb{Z}_q$-modules, i.e. of vector spaces over $\mathbb{Z}_q$. $\endgroup$ – anomaly Aug 7 '17 at 1:38
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    $\begingroup$ So the order is same as the order of GL$_{2}$($\mathbb{Z}_{q}$) i.e. (q$^{2}$-1)(q$^{2}$-q) ?? $\endgroup$ – HumbleStudent Aug 7 '17 at 1:43
  • $\begingroup$ Note proper MathJax usage, as in my edits to the question. $\endgroup$ – Michael Hardy Aug 7 '17 at 2:17
  • $\begingroup$ Wait, in saying $\mathbb{Z}_q$ are you referencing the additive group $(\mathbb{Z}/q\mathbb{Z})$ or the multiplicative group $(\mathbb{Z}/q\mathbb{Z})^*$? I'm assuming multiplicative since your limiting $q$ to prime numbers, correct? $\endgroup$ – Andrew Tawfeek Aug 7 '17 at 2:26
  • $\begingroup$ Correct, HumbleStudent. You get the group of invertible 2x2 matrices over $\Bbb{Z}_q$. $\endgroup$ – Jyrki Lahtonen Aug 7 '17 at 4:41

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