Determine what the automorphism group of $\Bbb Z_{7^3}$ looks like.

Question

I have a question about deciding what the automorphism group of $\Bbb Z_{7^3}$, that is ${\rm Aut}(\Bbb Z_{7^3})$, looks like.

I know that for a general prime $p$, the size of the group ${\rm Aut}(\Bbb Z_{p^3})$ equals the number of generators of $\Bbb Z_{p^3}$ since each automorphism is uniquely determined by the image of a generator.

Therefore, $|{\rm Aut}(\Bbb Z_{p^3})|$ = $p^2(p-1)$.

So for our case, we know that $|{\rm Aut}(\Bbb Z_{7^3})|$ = $7^2*6$. I got to here but I don't know how to continue.

How do I decide if ${\rm Aut}(\Bbb Z_{7^3})$ has a copy of $\Bbb Z_{7^2}$ or maybe two copies of $\Bbb Z_{7}$?

I think I know that it would have a copy of $\Bbb Z_6$ and not of $S_3$ because $\Bbb Z_{7^3}$ is cyclic, but I'm not quite sure...

Any help would be appreciated.

Answer 1

$\Bbb{Z}_{7^3}$ is a cyclic group so it's automorphism group is isomorphic to $\Bbb{Z}_{7^3}^{\times}$ (the set of multiplicatively invertible elements).

Answer 2

We have $${\rm Aut}(\Bbb Z_n)\cong U(n),$$ where $U(n)$ is the group of units modulo $n$.

Reference:

"Contemporary Abstract Algebra (Eighth Edition)," by J. A. Gallian, page 137.