Let $G$ be a group and $g \in G$. Show that $o(g)=o(xgx^{-1})$, where $o(x)$ is the order.
I've managed to show that the order of $xgx^{-1}$ is at most $o(g)$ by the following way.
Denote $o(g) = n$ and $o(xgx^{-1}) = m.$ Now $(xgx^{-1})^n=xgx^{-1} \cdots xgx^{-1} = xg^nx^{-1} = xx^{-1} = e.$
But I'm not sure how to show this the other way around? Any hints would be appreciated. This doesn't seem to yet imply that the orders would equal one another.
Using your argument you obtain that $m|n$.
Now observe that $g=x^{-1}(x g x^{-1}) x$ and using the same argument you obtain $n|m$.
It's much simpler than what you think: conjugation in a graoup is an automorphism, hence $$xg^nx^{-1}=e\iff g^n =e,$$ whence the orders of $xgx^{-1}$ and $g\:$ (least positive exponent such that $g^n=e$) are equal.
Lemma. For every $n\in\Bbb N$, $(xgx^{-1})^n=xg^nx^{-1}$.
Proof. Induction on $n$; trivially true for $n=1$ and true for $n$ (inductive hyp.); then: $$(xgx^{-1})^{n+1}=(xgx^{-1})^{n}xgx^{-1}\stackrel{(i.h.)}{=}xg^nx^{-1}xgx^{-1}=xg^{n+1}x^{-1}$$
$\Box$
Claim 1. $o(g)\mid o(xgx^{-1})$.
Proof. $e=(xgx^{-1})^{o(xgx^{-1})}\stackrel{(Lemma)}{=}xg^{o(xgx^{-1})}x^{-1}\Rightarrow g^{o(xgx^{-1})}=e\Rightarrow o(g)\mid o(xgx^{-1})$.
$\Box$
Claim 2. $o(xgx^{-1})\mid o(g)$.
Proof. $(xgx^{-1})^{o(g)}\stackrel{(Lemma)}{=}xg^{o(g)}x^{-1}=xx^{-1}=e\Rightarrow o(xgx^{-1})\mid o(g)$.
$\Box$
By the two claims, $o(g)=o(xgx^{-1})$.