I recently learnt in yet another post of mine how to get the ${\rm Aut}(\mathbb{Z}/4\mathbb{Z})$ i.e. $\mathbb{Z}/2\mathbb{Z}$, specifically consisting of the group $\{\alpha,\beta\}$ of these two automorphisms:$$\alpha: x\mapsto x, \beta: x\mapsto3x$$
These two automorphism can ben gotten also considering the automorphism group of the general cyclic group of order $n$ in yet another post.
So far, so good.
Now, my brain usually starts running freely asking usually to myself: what about other automorphism? Can there be any?
So I started reasoning on this further automorphism: $$\gamma: x\mapsto5x$$
Let's check this is an automorphism of $\mathbb{Z}/4\mathbb{Z} = \{0, \rho, 2\rho, 3\rho\}$
$$\gamma(e)=e$$$$\gamma(\rho)=5\rho=\rho$$$$\gamma(2\rho)=10\rho=2\rho$$$$\gamma(3\rho)=15\rho=3\rho$$
So, it's a legitimate automorphism sending $e\mapsto e$, since injective and also onto.
However, here this automorphism is manifestly behaving like the identity automorphism, i.e. $\alpha: x\mapsto x$. This can be seen also directly like this:
$$\gamma(x)=5x = 4x + x = x$$ , considering $4x=0$ for any $x$, since cyclic $\pmod{4}$.
So, if all the above is correct: am I right to claim $\gamma$ is all in all yet another identity automorphism, already accounted in $\{\alpha, \beta\}$?
Thanks in advance.
There are only two, right? $\rm{Aut}(\Bbb Z_4)\cong\Bbb Z_4^×\cong\Bbb Z_2$.
As I mentioned, since the residue class of $1$ in $\Bbb Z_4$ consists in $$\{\dots-7,-3,1,5,9,13,\dots \}$$ there's sort of infinitely many ways to write the identity automorphism.
Ditto the other one, where $1\to3$.