Suppose that $G$ is a transitive permutation group and suppose we know a construction of an isomorphism from the Galois group of a Galois number field to $G$. Does this information make it easier to solve the following more special inverse Galois problem:
Let $\tau$ be an involution of $G$ (i.e. $\tau^2$ is the identity), can we find complex Galois number field $K$ such that its Galois group is isomorphic to $G$ and such that this isomorphism identifies $\tau$ with the complex conjugation?
I was thinking in the lines of ... if I know that I can find an integral irreducible polynomial $f$ such that the splitting field of $f$ has a Galois group isomorphic to $G$ (from the fact that I know, constructively, that $G$ is Galois realizable) then maybe I can "magically" modify (aka. perturb the coefficients of) $f$ such that it remains irreducible, has complex roots, the Galois group remains isomorphic to $G$ and by some continuity argument we could "force" the complex conjugation to be $\tau$.
I'm too inexperienced in Galois inverse problems and maybe someone with more knowledge could give me some intuition.
Although there are some contexts in which one can do something along these lines, there is no known mechanism to do this in general. In fact, there are many groups for which the inverse Galois problem is known, but the "more restrictive" version where you insist that complex conjugation lies in some chosen conjugacy class is open. For example, if $G = \mathrm{GL}_2(\mathbf{F}_p)$, then there is a nice way to construct $G$-extensions by taking the $p$-torsion of a (general enough) elliptic curve $E/\mathbf{Q}$ and taking the corresponding number field. In this construction, complex conjugation is (for $p > 2$) always conjugate to
$$\left( \begin{matrix} 1 & 0 \\\ 0 & -1 \end{matrix} \right)$$
But if you now ask to find a $G$-extension in which complex conjugation is one of the other involutions (the identity or minus the identity) then this is an open problem (except for very small $p$).