Krdytkyu

$newcommand{gset}{Gtext{-}mathsf{set}}$
Let $G$ be any group and $gset$ be the category of $G$-sets, with morphisms being $G$-maps. That is an object of $gset$ is a pair $(X,rho)$ where $rho:Gtotext{aut}(X)$ is some group action of $G$ on $X$, and $$f:(X,rho)to (Y,tau),text{ satisfies } f(rho(g)x)=tau(g)f(x).$$

Am I correct that the fibred product $(X,rho)times_{f,(W,tau),g}(Y,chi)$ is given by $(Z,rhooplus chi)$ where
$$Z={(x,y)in Xtimes Ymid f(x)=g(y),$$
with the standard projections, and where by $rhooplus chi$ I mean that:
$$(rhooplus chi)(g)(x,y)=(rho(g)x,chi(g)y),$$
which makes the projection maps $G$-maps.

(And where by $f,g$ I mean that we are taking the fibred product with respect to the maps $f:(X,rho)to (W,tau)$ and $g:(Y,chi)to (W,tau)$.)

Further, are epimorphisms just set surjections that are $G$-maps?

That is correct. Conveniently, the forgetful functor $U:boldsymbol{G}mathbf{-set}tomathbf{Set}$ is monadic, so it creates limits; the upshot of which is that all limit constructions in $boldsymbol{G}mathbf{-set}$ have, as their underlying structure, the limit of the image of the same diagram under $U$ in $mathbf{Set}$. That doesn't tell you everything about the limit in $boldsymbol{G}mathbf{-set}$, but it's a good sanity check. That the set you describe has the requisite universal property takes very little checking.

And yes, epimorphisms are surjective $G$-maps. You can tell this is the case because the category of $G$-sets is equivalent to (if not defined as) the functor category $mathbf{Set}^G$, so $G$-maps are essentially just natural transformations, and a natural transformation of set-valued functors is epimorphic exactly when all of its components are epimorphic.