The question gives the "wrong" definition of Fix(T)$\operatorname{Fix}(T)$, hence the resulting confusion.
A more natural definition of the subfunctor X^G$X^G$ of "G"$G$-fixed points in X"$X$" is
(X^G)(T) = {x in X(T) | G_T-action on X_T fixes x}
= {x in X(T) | G(T')-action on X(T') fixes x for all T-schemes T'}.
$$
X^G(T) = \{x \in X(T) \mid\text{$G_T$-action on $X_T$ fixes $x$}\}
= \{x \in X(T) \mid\text{$G(T')$-action on $X(T') $ fixes $x$ for all $T$-schemes $T'$}\}.
$$
(Of course, can just as well restriction to affine T$T$ and T'$T'$ for "practical" purposes.)
By way of analogy with more classical situations, if the base is a field k then a moment's reflection with the case of finite k shows that
{x in X(k) | G(k) fixes x}
$\{x \in X(k) \mid\text{$ G(k)$ fixes $x$}\} $
is the "wrong" notion of (X^G)(k)$X^G(k)$, whereas
{x in X(k) | G-action on X fixes x}
is $
\{x \in X(k) \mid\text{$G$-action on $X$ fixes $x$}\}$ is a "better" notion, and is what the above definition of (X^G)(k)$X^G(k)$ says.
From this point of view, if (for simplicity of notation) the base scheme is an affine Spec(k)$\operatorname{Spec}(k)$ for a commutative ring k$k$ then the "scheme of G$G$-fixed points" exists whenever G$G$ is affine and X$X$ is separated provided that k[G]$k[G]$ is k$k$-free (or becomes so after faithfully flat extension on k$k$). So this works when k$k$ is a field, or any k$k$ if G$G$ is a k$k$-torus (or "of multiplicative type"). See Proposition A.8.10(1) in the book Pseudo-reductive groups.
