Forced conjugation of elements in finite groups
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.
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.
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.