Embedding of group such that two elements of same order become conjugate Fix the following objects: 
$G$= finite group, 
$x,y$ - distinct elements of $G$ of same order.
Q. Can we embded $G$ in a finite group $G_1$ such that $x,y$ become conjugate in $G_1$?

An application of HNN extension theorem perhaps does not consider embedding into finite groups for above problem (see Theorem 3.3 here). Perhaps, it simply done by adding a generator $z$ to $G$ and define a relation $z^{-1}xz=y$. But I couldn't ensure whether the embedding can be done in a finite group, provided original group is also finite.
 A: Actually, after thinking a bit more on it, this turns out to follow from the proof of Cayley's theorem. By construction, the embedding in that sends an element of order $m$ to a product of disjoint $m$-cycles (since this is the action by left translation), so all elements of the same order become conjugate.
A: The answer is yes. As Tobias Kildetoft indicated in the comments, in the regular representation of $G$ all non-identity elements act fixed point freely. So the two elements of the same order $n$ both consist of $|G|/n$ cycles of length $n$, and hence they are conjugate in the symmetric group on the elements of $G$.
More generally, if $A$ and $B$ are isomorphic subgroups of a finite group $G$ and $\phi:A \to B$ is an isomorphism, then you can embed $G$ in a finite group $X$ with an element $g \in X$ such that $g^{-1}ag=\phi(a)$ for all $a \in A$. This is a result of B.H. Neumann - I was looking for the original reference but I haven;t found it yet - he wrote several papers on similar themes.
