What are the all group automorphism of $Z_{60}$ sending 5 to 25?

Question

I am looking for all group automorphism of $Z_{60}$ sending 5 to 25.

I can see that all 16 generators are mapped to same elements and identity element is mapped to identity as well. Also that 30 is mapped to itself as it has order of 2. But there are still many elements unmapped.

Is there any other conditions that automorphism must satisfy?

Answer 1

Hint: $\Bbb Z_{60}\cong \Bbb Z_5\times\Bbb Z_{12}$, with $5\mapsto(0,5)$ and $25\mapsto (0,1)$.

Answer 2

You can use the chinese remainder theorem $\mathbb{Z}/60\mathbb{Z} \cong \mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/12\mathbb{Z}$ for example or bruteforce your way through the task by sending a chosen generator of $\mathbb{Z}/60\mathbb{Z}$ to another generator of $\mathbb{Z}/60\mathbb{Z}$. That means you should understand the generators. To get a grasp on all generators you have to consider all positive integers less than $60$ that are coprime to $60 = 2^2 \cdot 3 \cdot 5$.