Let $K=⟨\alpha⟩$ to be a cyclic finite group, my tutor stated that there is $f:K \rightarrow K$ such that $f(\alpha)=\alpha^k$, he said that it's a consequence from the statement bellow:
If $a$ also generates $K$ so there is a automorphism $f$ as $f(\alpha)=a$.
Then I think that $\alpha^k$ generates $K$ as well:
$⟨\alpha⟩=\{(\alpha^k)^n;\ n \in \mathbb{Z}\}=\{\alpha^{kn};\ n \in \mathbb{Z}\}$
since $kn \in \mathbb{Z}$
Is my point right?
By definition, the order of $\alpha^{k}$ is the least $m$ such that $km=o(\alpha)n$ for some $n$. Therefore, $m$ is of the form $\frac{o(\alpha)}{k/n}$ ($*$), and hence $o(\alpha^{k})$ ("...the least...") is gotten when the denominator in ($*$) is the greatest divisor of $k$ which is also a divisor of $o(\alpha)$: $$o(\alpha^k)=\frac{o(\alpha)}{\gcd(k,o(\alpha))}$$ So, in general, $\langle \alpha ^k\rangle $ is a proper subgroup of $\langle \alpha\rangle $, unless $k$ and $o(\alpha)$ are relatively prime.
(For example, think (additively!) of $G:=(\Bbb Z/6\Bbb Z,+)$: $1$ is a generator; which among $2\cdot 1=2$, $3\cdot 1=3$, $4\cdot 1=4$, $5\cdot 1=5$ are also generators of $G$?)