Let $G$ be a cyclic group, and $a,b$ two generators of $G$. Prove that exists automorphism $f$ such that $f(a)=b$ and that if $f$ is an automorphism then $f(a)$ generates $b$.
$G$ is cyclic, thus every $x\in G$ is $x=a^n$ and $x=b^m$. The considering $a^n= b^m$, is it possible to define $f(a)=a^{(n/m)}$? It'd be $f(a)=b$, but the $f$ defined needs to satisfy $f(ca)=f(a)$ if $f$ is automorphism, and with the $f$ I defined is $f(ca)=cf(a)$.
(1) define $f(a)=b$. Then take $f(a^k)=b^k$. That defines a bijection. To show that it is an automorphism all we need is that $f(gh)=f(g)f(h)$. But that follows immediately from the definition of $f$.
(2) Here we are given that $f$ is an automorphism. So suppose $f(a)=g$. We want to show that $g^k=b$ for some $k$. But $f$ must be surjective, so if the order of the group is $n$, then the elements $f(a),f(a^2),\dots f(a^n)$ must be just a rearrangement of the elements of $G$, so we must have $f(a^k)=b$ for some $k$. But $f(a^k)=g^k$, so $g^k=b$.