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. 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 $k$.