Let $X$ a (locally Noetherian; but not sure if that's really matter) $k$- scheme, $G$ a $k$-group scheme acting on $X$ via morphism $a:X \times G \to X$.

The *fixed point functor* of $X$ (where $X$ is regarder as usual as point functor associating to a $k$-scheme $S$ the set $X(S)=\text{Hom}(S,X)$ of $S$-valued points) is the subfunctor $X^G$ that associates with any scheme $S$, the set $X^G(S)$ of all $x \in X(S)$ such that for any $S$-scheme $S'$   and any $g \in G(S')$, we have $g \cdot x = x$ (as element inside $X(S')$.  
(The definition is taken from [these notes][1] by M. Brion.)


**Two (maybe two and half) questions:**  
**(1):** What is the precise reason to declare the fixed point functor to be "stable" with respect to $S$-extensions $S'$ as defined above?  
Asking differently, what would go wrong (caused precisely due to which phenomena?)  if we would try to define it "naively", namely $X^G(S)$ as the set of all $x \in X(S)$ such that for any $g \in G(S)$, we have $g \cdot x = x$ (as element inside $X(S)$, ie without taking $S$-extensions $S'$ into account?  

Which potential problems would we facing then? Probably the most plausible formal reason coming into my mind might be that the subfunctor defined that way could happen to be not representable anymore, right?  
(Note it's a *fact* that if a subfunctor representable by a scheme it neccessarily a fppf sheaf / satisfies fppf descent, so it's plausible (...and in light of this fact even neccessary) that we should demand it to be stable wrt certain $S$- extensions $S'$.  

But if that's really the *only* distinguishing reason to define the fixed point functor as in the notes and not in "naive" approach as I did above, why it is formulated in terms of *all* $S$- extensions $S'$ and not only the fppf ones?  

**(2):** The notes refer for explicit construction of scheme representing $X^G$ to Demazure's & Gabriel's "Groupes algebriques" or SGA, where in that full generality in the construction is a lot involved.

My question is if the construction can be "simplified" (especially the "geomertic" picture becomes more plausible) if we add additional assumption that $X$ ( and so $X^S$) contains a $k$- rational point fixed by $G$-action? 

(Motivation: In algebraic geometry there are a lot of examples for constructions defined on level of arbitrary base $k$ ( or even $S$- based schemes) which became  much more "geometrical" if we additionally assume the existence of *rational* points (e.g.: a connected $k$-scheme which has a $k$- rational point is already geometrically connected).  
So my hope was that if we would allow here the existence of a $k$- rational point $x \in X(k)$ fixed by $G$, then the construction of $X^G$ may simplify, eg relizable as certain fiber product, instead of be given through rather involved constructions as in the refered sources still working under much weaker assumptions.



[1]: https://www-fourier.ujf-grenoble.fr/~mbrion/course_rev.pdf "Some structure theorems for algebraic groups"