Forced conjugation of elements in finite groups The dihedral group
$$
D_8 = \langle\ a,\ b \ \mid \ a^4,\ b^2,\ (ab)^2\ \rangle
$$
has a central involution $c=a^2$ and a non-central one, $b$. 

Q. Can we embed $D_8$ into a finite group $G$ in which $c$ and $b$ become conjugate?

 A: Regarding your specific example, the dihedral group of order $8$ admits an embedding into $S_4$, namely $$D_8 \simeq \langle (1 \, 2 \, 3 \, 4), \, (1 \, 3) \rangle.$$
The center is generated by $(1 \, 3)(2 \, 4)$, and the remaining involutions are $(2 \, 4)$, $(1\, 2)(3 \,4)$, $(1 \, 4)(2 \, 3)$, $(1 \, 3)$. 
Then in $S_4$ the central involution of $D_4$ becomes conjugate to two non-central ones.  
A: Actually for any finite group $H$, any two elements of the same order are conjugate in some larger finite group $G$.
Proof: Consider the embedding into the symmetric group given by the permutation action on elements of $H$ by left multiplication. An element of order $n$ has $|H|/n$ orbits, all of size $n$, which uniquely determines the conjugacy class.
A: Will Sawin's lovely answer deals with the question in much greater generality, but for the specific case in hand, many Finite Group Theory texts ( eg Gorenstein (1968)) use groups with dihedral Sylow $2$-subgroups to illustrate possible fusion patterns via Alperin's fusion theorem. I will discuss the possibilities in the case of a finite group $G$ with a dihedral Sylow $2$-subgroup $D$ of order $8$ (ie with $8$ elements). There are three different possibilities. Note that $D$ has two different Klein $4$-subgroups $U$ and $V$, and that since ${\rm Aut}(D)$ is a $2$-group, we have $G = DC_{G}(D).$ By a Theorem of Burnside, $U$ and $V$ are not conjugate in $G$ (since they are certainly not conjugate in $N_{G}(D)$ and both are normal in $D).$ The three possibilities are:


*

*$N_{G}(U)/C_{G}(U) \cong N_{G}(V)/C_{G}(V) \cong \mathbb{Z}/2\mathbb{Z}.$ In this case, $G$ has a normal $2$-complement, and involutions of $D$ are conjugate in $G$ if and only if they are already conjugate in $D.$

*$N_{G}(U)/C_{G}(U) \cong \mathbb{Z}/2\mathbb{Z}$ and $N_{G}(V)/C_{G}(V) \cong S_{3}.$ In this case, $G$ does not have a normal $2$-complement but does have a normal subgroup of index $2.$ In this case, all the involutions of $V$ all become conjugate in $G,$ but not all involutions of $U$ are $G$-conjugate. The example of $G = S_{4}$ given by Francesco Polizzi gives an example where this occurs.

*$N_{G}(U)/C_{G}(U) \cong N_{G}(V)/C_{G}(V) \cong S_{3}.$ In this case, $G$ has no factor group of order $2$ and all $5$ involutions of $D$ are conjugate in $G$. The examples of $G \cong  A_{6}$ or $G \cong {\rm PSL}(2,7)$ are cases where this occurs (there are many more examples, of course).
Later edit: regarding the exchange in comments between @YCor and @Derek Holt : it is clear from this analysis that $S_{4}$ is the smallest group with a subgroup $D_{8}$ such that a non-central involution of $D_{8}$ and a central involution of $D_{8}$ becoming conjugate in the overgroup. Also ${\rm PSL}(2,7)$ is the smallest group with a subgroup $D_{8}$ whose involutions are all conjugate in the overgroup.
A: (1) (essentially HJRW's comment to Will Sawin's answer) 
Given a finite group $F$ and an isomorphism $t$ between two subgroups $A,B$ of $F$, the universal group in which $A$ and $B$ are conjugate through $t$, namely the HNN extension $H$ of $(F,A,B,t)$, 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 $t$). This applies in particular to the case when $A,B$ are cyclic subgroups of the same order.
Of course this argument, which looks somewhat immediate at first glance, is less elementary than the one given by Will Sawin since it relies on residual finiteness of HNN extensions of finite groups. 
(2) Will's argument extends to this setting ($F$, $A$, $B$ without assuming $A,B$ cyclic. Namely, $A$ 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 $\sigma$.
