Trying to find an isomorphism

Question

I'm trying to find an isomorphism from Z31 to itself that maps 13 to 28.
I used the extended euclid algorithm to find that 12 is the inverse of 13, but that's as far as I got. What do I do next? How can I find this specific isomorphism?

Answer 1

The morphisms from $\mathbb Z_n$ to $\mathbb Z_n$ consist of picking an integer $a$ and sending $x$ to $ax$. (in fact the morphisms from any cyclic group $\langle a \rangle$ satisfy $f(na)=nf(a)$).

We just need to find an $a$ so that $13a\equiv 28$, since $13^{1}=12$ your $a$ is $28\times12\equiv 26$.

So your isomorphism is $f(x)=26x$. Notice it is injective since $(26,31)=1$. since the groups are finite it is an isomorphism.

Answer 2

If you define an automorphism $\varphi$ of $\mathbf Z_{31}$ by the image $a$ of $1$, you must have $$\varphi(13)=13\,\varphi(1)=13a =28,$$ whence $$a=13^{-1}28=12\cdot 28=26.$$