Dualizing the morphic group action Let $G$ be an algebraic group and $X$ an algebraic variety. I am confused about dualising the morphic action of an algebraic group on a variety. So we have $\psi:G\times X\to X$ is a group action. Then we can consider $\psi_g:X\to X$ for each $g\in G$ if we wish. This feels very natural to me. Now when we dualize, we obtain $$(1):\quad \psi^*:K[X]\to K[G]\otimes K[X].$$ So to dualize, I am of the maybe false opinion, we are essentially working with a pullback. Say we were working with the left translation $\psi_x(y)=x^{-1}y$. Then we can define on $f\in K[X]$ $f(x^{-1}y)=f\circ \psi_x(y)=\psi_x^*(f)(y)$, so we can naturally think of $\psi_x^*:K[X]\to K[X]$. But then, moving to $\psi^*$, surely I obtain: $$\psi^*:G\times K[X]\to K[X],$$ rather than obtaining $(1)$. I guess this is remedied by not using the pullback property. But then, what do we want to capture with this dual map?

If $G$ is any group acting on an algebraic variety $X$, we get an action of $G$ on $K[X]$ (ie a map $G\times K[X]\to K[X]$) as you describe. However, this doesn't say anything about whether the action respects the algebraic structure of $G$ (if it has one). To make a geometric analogy, your proposed construction is like having an action of a Lie group on a manifold such that $gm$ is a continuous function of $m$ but maybe not a continuous function of $g$.

A ring homomorphism as in (1) is like a group action that is continuous in both variables. You can recover your maps $\psi_x^*$ if you like. For each closed point $g\in G$ we have a map $e_g:K[G]\to K$ (namely evaluation at $g$), and composing with $\psi^*$ gives a map $$ (e_g\otimes1)\psi^*:K[X]\to K[X]. $$ This is your $\psi_{g^{-1}}^*$ (the inverse is just to keep it a left action).