what are the other 2 nontrivial elements of the automorphism group of $\Bbb Z/5\Bbb Z$?

Question

It is known that the automorphism group of the units of $\Bbb Z/5\Bbb Z$ is isomorphic to the cyclic group of order $4$, so the automorphism group must also have $4$ elements. The two nontrivial ones are the identity and the conjugation map, what are the other $2$?

Answer 1

Rats, I misread the original question. You're looking for $\text{Aut}(\Bbb Z/5\Bbb Z)$, which is indeed isomorphic to $\Bbb Z/4\Bbb Z$. Since this group is cyclic, it's enough to find an element of order $4$, and then all choices are given by powers of that generator.

Indeed, multiplication by $-1$ gives you inversion, which is an element of $\text{Aut}(\Bbb Z/5\Bbb Z)$ of order $2$, so what should give us that upon using it twice? The answer is: multiplication by $i=\sqrt{-1}$! But what would that mean in $\Bbb Z/5\Bbb Z$? Well, here we see that $2^2= 4\equiv -1\mod 5$, so $2$ acts like a square root of $-1\mod 5$, in fact in the field $\Bbb F_5$, $2$ is a primitive fourth root of unity, so it behaves algebraically just like $i$.

So all automorphisms are given by:

$$\varphi_n([k])=[2^nk]\mod 5 =[i^nk]\mod 5, \; 0\le n\le 3$$

You can check that these are group isomorphisms directly:

(1) Since $\gcd(2,5)=1$ there exist $x,y$ such that $2x+5y=1$ by the Euclidean algorithm. But then, the map $[n]\mapsto [2n]$ is injective since it has inverse $[m]\mapsto [xm]$, and since the domain and range have the same (finite) size, it is a bijection. (2) by the distributive property

$$\varphi_n([k]+[j])=[2^n(k+j)]=[2^nk+2^nj]=[2^nk]+[2^nj]=\varphi_n([k])+\varphi_n([j])$$

so that you have a group homomorphism for each $n$.

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Note: This fully generalizes to other prime order fields. Each non-zero element there is a $p-1^{st}$ root of unity (Fermat's theorem), and finding a primitive one, $\zeta\in\Bbb Z/p\Bbb Z$ gives you the generator for the automorphism group, namely

$$\varphi_\zeta([k])=[\zeta k]$$

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Addendum

Based on an update from the op, this is suppsed to be a Galois group for the field $\Bbb Q(\zeta_5)/\Bbb Q$, luckily such a thing is determined on its action on a generator, which is a $5$th root of 1. The operative thing is that a generator permute the $5$th roots of $1$ among themselves and have distinct maps for all of its iterates, $f, f^2, f^3, f^4$ where the power is functional composition.

But then

$$f(\zeta_5)=\zeta^2$$

is just the map, for $f^k(\zeta_5)=\zeta_5^{2^k}$ and since $\{2^k: 1\le k\le 4\}$ generates the non-zero residue classes modulo $5$, by our previous discussion, it is seen that the automoprhism $f$ has order $4$, hence generates the Galois group.