TITLE: Given two elements $g,h$ in a finite group $G$ of the same order, does there exist a finite group $G'$ containing $G$ where they are conjugate?
QUESTION [8 upvotes]: Let $G$ be a finite group, and let $g,h\in G$ be two elements of the same order. Does there exist a larger finite group $G'\ge G$ such that $g,h$ are conjugate in $G'$?
If not, what is known about groups which have this property?

REPLY [2 votes]: Since the question was reopened in spite of a duplicate in MathSE, I also "duplicate" my answer from there (a little improved anyway), so as to provide a self-contained answer. I leave it cw since it relies on ideas form other answers and comments.
(a) The answer to the OP's question is yes, and more generally one has:

Fact. Consider a finite group $F$ and two subgroups $A,B$ with an isomorphism $t:A\to B$. Then one can embed $F$ in a larger finite group in which $A$ and $B$ are conjugate (the conjugation inducing the isomorphism $t$ from $A$ to $B$).

This applies in particular to the case when $A,B$ are cyclic subgroups of the same order.
Proof of the fact. $A$ has two free actions on $F$, one being given by $a\cdot g=ag$, the other by $a\cdot g=t(a)g$. Writing $k=|F/A|$, find $k$ points $x_1,\dots,x_k$, one in each orbit of the first action, and $k$ points $y_1,\dots,y_k$, one in each orbit of the second action. Then extend the assignment $x_i\mapsto y_i$ to a permutation $\sigma$ of $F$, by the assignment $\sigma(ax_i)=t(a)y_i$. This is well-defined by freeness of the action. Then $$\sigma(abx_i)=t(ab)y_i=t(a)t(b)y_i=t(a)\sigma(bx_i)$$ for all $a,b\in A$ and $i$, and thus $\sigma(ag)=t(a)\sigma(g)$ for all $a\in A$ and $g\in F$. In other words, $\sigma\circ L_a=L_{t(a)}\circ \sigma$. So in the permutation group of $F$, where $F$ is identified to its image through left multiplication, the isomorphism $t:A\to B$ is realized by conjugation by $\sigma$. $\Box$
This is essentially Will's argument, which was only written for $A,B$ cyclic.
(b) In the same setting, the universal group in which $A$ and $B$ are conjugate through $t$, is the HNN extension $H$ of $(F,A,B,t)$, defined by the relative presentation
$$\langle u,G\mid uau^{-1}=t(a),\forall a\in A\rangle.$$
It is well-known that this HNN-extension is virtually free and hence residually finite. Thus, there exists a finite quotient of $H$ in which $F$ is mapped injectively, and in this quotient $A$ and $B$ are indeed conjugate (by $t$). 
Of course this argument (mentioned by HJRW in the comments to Will's answer), which looks somewhat immediate at first glance, is less elementary than the one in (a) since it relies on residual finiteness of HNN extensions of finite groups; anyway it's a very natural proof since the construction is universal.