Let $(G, \cdot)$ be a group and let $a \in G$ be a finite ordered element and let $n$ be an arbitrary integer. Prove that $a^n$ is also a finite ordered element. Prove that $a$ and $a^n$ have equal orders if and only if $\gcd\big(n, o(a)\big) = 1$.
I managed to prove the first part, but didn't manage to prove the second part.
Proof of first part: Observe that $\langle a^n\rangle = \{(a^n)^p \ | \ p \in \mathbb{Z}\} = \{(a^{n\cdot p}) \ | \ p \in \mathbb{Z}\} \subseteq \langle a \rangle = \{a^m \ | m \in \mathbb{Z}\}$ and since $o(a)$ is finite we have $\langle a \rangle$ to be finite and thus $\langle a^n \rangle $ to be finite. Since $o(a^n)$ is the cardinilatity of the set $\langle a^n \rangle $ we have $o(a^n)$ to be finite and thus $a^n$ is also a finite ordered element. $\square$
Firstly, is my proof above satisfactory? Secondly I am not sure how to go about proving the second part of the problem, how could one prove the second part of this problem?
Your proof looks correct, but there is perhaps an easier way:
Since $o(a)$ is finite, then $(a^{n})^{o(a)}=e$, so $o(a^n)\mid o(a)n$ and $o(a^n)$ is therefore also finite.
For the second part try proving that if $\gcd(o(a),n)=1$ then $o(a)\mid o(a^n)$ and $o(a^n)\mid o(a)$. This will prove that $o(a)=o(a^n)$. Here is a hint, notice that $$a^{o(a^n)n}=e=a^{o(a)n}$$
For the other direction you assume that $$a^{o(a)}=(a^n)^{o(a)}$$ and need to show that $\gcd(o(a),n)=1$. Remember that $o(a)$ is the smallest number such that $a^{o(a)}=e$, what does this tell you about $o(a)n$?