How many isomorphism of $\phi :\mathbb Z_{4} \rightarrow \mathbb Z_{4}$? I'm interested in how to find it, not the answer itself. I'm confuse to solve this question, I know isomorphism is bijective, and in this case it called Automorphism. But, I can't find a way how to find the answer.
 A: Notice that an automorphism $f$ is entirely determined by $f(1)$ which's a generator of $\mathbb Z_4$ so we have two possibilities:


*

*$f(1)=1$ and in this case $f=\operatorname{id}$

*$f(1)=3=-1$ and in this case $f=-\operatorname{id}$.

A: Note that $\mathbb{Z}_4$ is cyclic group, so the number of automorphisms can be found using the euler-totient function:
$$\varphi(4)=2$$
Thus there are 2 automorphisms of $\mathbb{Z}_4$. The isomorphisms are:
$$\theta(1)=1$$
$$\theta(1)=3$$
A: Any group homomorphism $\pi:\mathbb Z_4\to G$ is determined by the image of $1$, since $\mathbb Z_4$ is a cyclic group generated by 1, so $\pi(k)=\pi(1+\dots+1)=\pi(1)+\dots+\pi(1)$. Since isomorphisms preserve order, the generator $1$ (of order 4) has to be mapped to another element of order 4. The only elements of order 4 in $\mathbb Z_4$ are $1$ and $3$, so those are the only valid choices to obtain an isomorphism. Choose $1$ and you get the identity automorphism, choose $3$ and you get the automorphism $k\mapsto -k$.
A: Since $\mathbb Z_4$ is cyclic with generator $1$ or $3$, every homomorphism $\varphi$ from $\mathbb Z_4$ is completely determined by $\varphi(1)$ or $\varphi(3)$. Let's pick $1$ to work on.
If $\varphi(1) = 0$, then $\varphi$ is the zero map, hence not an automorphism.
If $\varphi(1) = 1$, we have an identity map, which is obviously an automorphism.
If $\varphi(1) = 2$, then $\varphi(2) = \varphi(1) + \varphi(1) = 2 + 2 = 0 = \varphi(0)$. This map is not surjective, hence not an automorphism.
If $\varphi(1) = 3$, then $\varphi(2) = 3 + 3 = 2$ and $\varphi(3) = 3 + 3 + 3 = 1$. This is also an automorphism.
We have two automorphisms.
