Automorphism intuition

Question

I get why if $(\mathbb{Z}/3,+)$ is our group we have $\text{Aut}(\mathbb{Z}/3) \simeq \mathbb{Z}/2$ since a group $G$ where $|G|=p$ is isomorphic to $\mathbb{Z_p}$ where $p$ is a prime, but what is the intuition behind this? $\text{Aut}(\mathbb{Z}/3)$ is a group of isomorophism mappings from $\mathbb{Z}/3$ to $\mathbb{Z}/3$ and $\mathbb{Z}/2$ is a group with elements from the set of integers, $\{0,1\}$ so how can a group of isomorphism mappings be isomorphic (intuitively an isomorphism means the same group but with elements just labeled differently) to a group with elements from a set? I don't see how isomorphic mappings are equal to integers.

Answer 1

Let $\varphi: \mathbf{Z}/3 \to \mathbf{Z}/3$. Any homomorphism between cyclic groups is fixed by considering the image of a generator, so let's just ask ourselves where it sends $1$. There are three possibilities: it can go to $0$, $1$, or $2$. Now let's impose the condition of being an automorphism. You'll see quickly that it cannot go to $0$, otherwise it would not be injective. There are no other restrictions.

You can generalize this reasoning to deduce that

For a cyclic group of order $p$, where $p > 2$ is prime, then its automorphism group is cyclic of order $p-1$.

Answer 2

Identify $\DeclareMathOperator\aut{Aut}\aut(\mathbf Z/3\mathbf Z)$ with the images of $1$ in $\mathbf Z/3\mathbf Z$, which may be $1$ or $-1$ and the additive group $\mathbf Z/2\mathbf Z$ with the multiplicative group $\bigl\{\pm 1\bigr\}$.

More generally, $\aut(\mathbf Z/p\mathbf Z)$ can be identified with the multiplicative group of units $\bigl(\mathbf Z/p\mathbf Z\bigr)^\times$, which is a cyclic group isomorphic to the additive group $\mathbf Z/(p-1)\mathbf Z$.