Does there exists group isomorphism $ϕ: \Bbb Z_{40} \rightarrow \Bbb Z_{40}$ such that $ϕ(17) = 15$?

Question

Does there exists group isomorphism $ϕ: \Bbb Z_{40} \rightarrow \Bbb Z_{40}$ such that $ϕ(17) = 15$?

What I tried so far: $ϕ(17) = 17*ϕ(1) = 15$.

And now I wanted to show the equation $17x = 15 \pmod {40}$ have no solution, But this didn't got me anywhere.

Any hints will be usefull :)

Answer 1

$\gcd(17,40)=1$. So $\overline{17}$ generates the group $\mathbb{Z}_{40}$.

The isomorphic image of this generator $\overline{17}$ must also be a generator. However $15$ does not generate the group.

$17x=15$ will have a solution because $17$ is a generator of the group.

In general if you have a cyclic group $G=\langle a\rangle$. Then $f:G\to G$ is an isomorphism(called Automorphism) iff $G=\langle f(a)\rangle$.