How can I conclude that an automorphism is the identity map?

Question

Let $\phi$ be an automorphism on a group $G$. If $\phi$ maps any one non-identity element in $G$ to itself, is $\phi$ necessarily the identity map? What if $G$ is cyclic?

Answer 1

What about the cyclic group $Z/8Z$? Negation is the automorphism taking x to −x. It leaves both 4 and 0 fixed. Therefore mapping a non-identity element to itself is not enough even in the case of cyclic groups.

Answer 2

Let $H$ be any group and $G = H\times H$. Then $$\phi : G\to G, \quad (x,y) \mapsto (y,x)$$ is an automorphism of $G$. The elements $(x,x)$ are fixed under $\phi$. As long as $H$ is not the trivial group, $\phi\neq\operatorname{id}_G$ and there are always non-identity elements of the form $(x,x)$.

Answer 3

If $G$ is cyclic then it is isomorphic to $\mathbb Z_n$ for $n\in\mathbb N$ or to $\mathbb Z$. The automorphism group on $\mathrm{Aut}(\mathbb Z_n)$ is isomorphic to $\mathbb Z_n^\times$ which is the multiplicative group of integers mod $n$ that are coprime to $n$. A $k\in\mathbb Z_n^\times$ corresponds to $\phi_k$ which maps the generator $1$ to $k$. If the map has a fixed point $l$, this implies $\exists m\in\mathbb N: kl=mn+l$, thus $n|(k-1)l$. Since $k-1<n$, $l$ has to satisfy $\mathrm{gcd}(n,l)>1$.

This means that an automorphism on $\mathbb Z_p$ for $p$ prime has no fixed point, whereas for arbitrary $n$ there is likely to be an automorphism with a fixed point.
For example, $n$ is always mapped to itself by every automorphism on $\mathbb Z_{2n}$.

On the other hand $\mathbb Z$ has only two automorphisms: identity and negation.