Let $G$ be an algebraic group acting on $X$ (a finite type scheme on $k$).


A $G$-invariant $k$-morphism $f : X \rightarrow S$ is a map such that the following commute:
$\require{AMScd}$
\begin{CD}
G \times\_k X @>\rho>> X\\
@V \pi\_2 V V @VV f V\\
X @>>f> S
\end{CD}


Where $\rho$ is the action map and $\pi\_2$ is the projection on the second coordinate.


Every $k$-point $g \in G(k)$ induces a map $\phi\_g : X \rightarrow X$ given by the composition:
$\require{AMScd}$
\begin{CD}
 X = \mathop{Spec} k \times\_k X @>{(g, Id)}>> G \times\_k X @>{\rho}>> X 
\end{CD}
We now define a second kind of invariance: $f : X \rightarrow S$ is invariant if $f= f \circ \phi\_g$ for all $g \in G(k)$.


Obviously, the first definition implies the second one. **Is it true the inverse implication?**


In my case $k$ is algebraically closed, but I don't know if I need more assumptions. If yes, do you know some counterexamples?
It is true if I suppose $X$ and $G$ reduced, but I hope I can avoid this.


Furtermore: Let's consider the map induced on $k$-points: $ \rho(k) : G(k) \times X(k) \rightarrow X(k)$. Let $k$ be algebraically closed and let $X$, $G$ be reduced. It's well known that $\rho(k)$ determines $\rho$. It is true if I do not assume $G$ to be reduced?


**My real question:**
My question comes from coarse moduli space, in particular from proposition 3.35 of [these notes](https://userpage.fu-berlin.de/hoskins/M15_Lecture_notes.pdf). It isn't clear to me why $\eta\_S(\mathcal{F})$ is $G$-invariant (with reference to pdf notations). The proposition doesn't request that $X$ and/or $G$ are reduced, but it seems that the proof assumes it.


An algebraic group is a group scheme of finite type over $k$.
Every scheme is intended to be of finite type over $k$.



$G$-invariant morphism and coarse moduli spaces
This is not true without the assumption that $G$ is reduced. Here is a counterexample.


Fix a prime number $p$ and any field $k$ of characteristic $p$. We define the nonreduced algebraic subgroup $\alpha\_p \subset \mathbb{G}\_a$ by $\alpha\_p = Spec(k[t]/(t^p))$ (here $\mathbb{G}\_a$ denotes the additive group over $k$). Then $\alpha\_p$ contains a single $k$-point-- the identity $0$ of the group.


We set $X = \mathbb{G}\_a$. The group $\alpha\_p$ acts on $X$ by addition (include $\alpha\_p$ into $\mathbb{G}\_a$ and then use the group structure).


The identity morphism $id: X \to X$ is not $\alpha\_p$-invariant (the scheme-theoretic image of $\alpha\_p \times 0 \subset \alpha\_p \times X$ is just $0\subset X$ under the projection and $\alpha\_p \subset X$ under the action). However, it is clearly invariant under the unique $k$-point of $\alpha\_p$ (the identity of the group).


If $G$ is reduced and $S$ is separated, then I think that you should be fine (under your assumption that $k$ is algebraically closed).


Edited:


For the last part, you can argue as follows. Consider the morphism
$$ \bigsqcup\_{p \in G(k)} p \to G$$
By the Nullstellensatz, this has schematic image $G$. Using an argument similar to the end of the proof of Prop 3.2.4 (ii) in [http://virtualmath1.stanford.edu/~conrad/249BW16Page/handouts/alggroups.pdf](http://virtualmath1.stanford.edu/%7Econrad/249BW16Page/handouts/alggroups.pdf), you can show by arguing affine locally that the product morphism
$$\bigsqcup\_{p \in G(k)} p \times X \to G \times X$$
has schematic image $G \times X$. (Notice that the original morphism is not quasicompact! so it is not automatic that the schematic image commutes with flat base-change, but in this case it does. This is true when your base is a field because any ring over $k$ is a free $k$-module).


Now, if $S$ is separated, the locus of $G \times X$ where the projection agrees with the action is a closed subscheme $Z \subset G \times X$. By assumption, we have a factorization
$$\bigsqcup\_{p \in G(k)} p \times X \to Z \to G \times X$$
Therefore, by the scheme theoretic density we must have $Z = G \times X$.


