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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 regarded as usual as point functor associating to a $k$-scheme $S$ the set $X(S)=\operatorname{Hom}(S,X)$ 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 by M. Brion.)

Two (maybe two and half) questions:

Question 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 effects?) if we would try to define it "naïvely", 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 $X(S)$, i.e., 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 that it's a fact that if a subfunctor is representable by a scheme it is necessarily a fppf sheaf / satisfies fppf descent, so it's plausible (…and in light of this fact even necessary) that we should demand that it has 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 "naïve" approach as I did above, why it is formulated in terms of all $S$-extensions $S'$ and not only the fppf ones?

#EDIT: Meanwhile Q1 got resolved thanks to Alex Youcis; see comments. The issue is that with the "naïve" definition as I proposed above it as Alex' example shows would not even be a subfunctor.)

Question 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 "geometric" picture becomes more imaginable) if we add the additional assumption that $X$ contains a $k$-rational point fixed by $G$-action?

(Motivation: In algebraic geometry there are a lot of examples for constructions defined in broad generality ("EGA style") on level of arbitrary base $k$ (or even $S$-based schemes for arbitrary scheme $S$) 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, e.g., be realizable immediately as a certain fiber product, instead of be given through "multiple staged" rather involved construction as in the referred sources still working under much weaker assumptions.)

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    $\begingroup$ Is it even functorial then? For instance, consider $\mu_p$ acting on $\mathbb{A}^1_{\mathbb{F}_p}$ by $g\cdot x=gx$. Then, for your naive fixed points we would get over $\mathbb{F}_p$ the set $\{x\in \mathbb{F}_p:gx=x\text{ for all }g\in \mu_p(\mathbb{F}_p)\}$. But, $\mu_p(\mathbb{F}_p)$ is trivial, so this is just $\mathbb{F}_p$. But, if we plug in $\mathbb{F}_p[T]/(T^p-1)$ then we would get $\{x\in\mathbb{F}_p[T]/(T^p-1): gx=x\text{ for all }g\in \mu_p(\mathbb{F}_p[T]/(T^p-1)\}$. In particular, it's not true that $1\in\mathbb{F}_p$ is in here as $T\cdot 1=T\ne 1$. $\endgroup$
    – Alex Youcis
    Commented May 8 at 10:07
  • $\begingroup$ @AlexYoucis: oh right, I see, so seemingly this "naive" definition would already fail to declare it as well defined functor, thanks! $\endgroup$
    – user267839
    Commented May 8 at 10:21
  • $\begingroup$ @AlexYoucis, re, we can even see the problem purely on the level of field points, replacing $\mathbb F_p$ and $p$ by $\mathbb Q$ and $3$. \\ @‍OP, I am having trouble parsing "Note it's a fact that if a subfunctor representable by a scheme it necessarily a fppf sheaf / satisfies fppf descent". At first I thought that the last "it" should be "is", but that doesn't read right either to me (what is the consequent of the "if"?). Could you fix it? $\endgroup$
    – LSpice
    Commented May 8 at 15:09
  • $\begingroup$ @LSpice: sure, sorry for pretty poor wording $\endgroup$
    – user267839
    Commented May 8 at 15:14

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