We say that group $G$ is conjugacy separable if for every $g \in G$ the set $g^G = \{cgc^{-1} \mid c \in G\}$ is closed in the profinite topology on $G$, i.e. for every $f \in G \setminus g^G$ there is a finite index normal subgroup $N$ such that $fN \cup g^G = \emptyset$.
Now suppose that $G = K \rtimes R$ is a conjugacy separable group. One can easily show that for every $r \in R$ the set $r^R = \{crc^{-1}\mid c \in R\}$ is closed in the profinite topology on $G$. Can we show the same for elements $k \in K$: is the set $k^R$ closed in $G$ for every $k\in K$? If not, what would be an example of conjugacy separable group that splits as a nontrivial semidirect product and a tuple of elements that cannot be separated?
Here is a requested example. Let $K=\langle a,t \mid tat^{-1}=a^2\rangle$ be the Baumslag-Solitar group $BS(1,2)$, and let $G=K \times \langle b \rangle_\infty$ be the direct product of $K$ with an infinite cyclic group. Then $G$ is conjugacy separable as a direct product of c.s. groups (conjugacy separability of K is well-known, and in a recent survey of Moldavanskii is attributed to [Moldavanskii D. I., Kravchenko L. V., Frolova E. N. "Conjugacy separability of some one-relator groups", Algorithmic Problems in Group and Semigroup Theory. Tula Pedagogical Inst. 1986. pp. 81–91 (Russian)]).
Now, let $R \leqslant G$ be the infinite cyclic subgroup generated by the pair $(a,b)$. Clearly $G\cong K \rtimes R$, and the conjugacy class $(t,1)^R$ is the subset $\{(a^nt,1) \mid n \in \mathbb Z\}=\langle (a,1) \rangle (t,1)$. The subgroup $\langle(a,1)\rangle$ is conjugate to its proper subgroup $\langle (a^2,1)\rangle$ in $G$, hence it cannot be closed in the profinite topology, so its translation $(t,1)^R$ will not be closed either.