Points on base-change of group scheme Let $K = k(t), S = \mathrm{Spec}(K)$ and let $\phi : G \to S$ be an $S$-group scheme. Define the $k$-algebra morphism $f_0 : K \to K$ by $f(t) = t^m$ (for some integer $m>0$). This induces a morphism $f : S \to S$.
Let $G' := G \times_{\phi, f} S$ be the fiber product over $S$ using the morphisms $\phi, f : G, S \to S$. This is a group scheme over $S$.

I want to prove that $G'(k(t)) = G'(K) := G'(S)$ is isomorphic (as a group) to $G(k(t^{1/m}))$.


This seems intuitive since $G'$ is obtained from $G$ by replacing $t$ by $t^m$ in all the local equations. Now, the $S$-points of $G'$ are
$$G'(S) \simeq G(S) \times_{S(S)} S(S)$$
But for any $S$-scheme, the $S$-points of $S$ are a singleton: $\mathrm{Hom}_S(S, S) = \{ id_S \}$, so I am confused because it seems that I get a bijection $G'(S) \simeq G(S)$... (Actually I have the same problem understanding the points of the relative Frobenius $G^{(p)}$ when $k$ has characteristic $p>0$).
 A: For me, the easiest way to think about base changing objects in algebraic geometry is via the 'functor of points' viewpoint.
Given a ring $R$ and an $R$-scheme $X$ we can think of $X$ as a functor $\mathsf{Alg}_R \to \mathsf{Set}$.
Now given a ring homomorphism $\varphi \colon R \to R'$, the base change $X \otimes_R R'$ (viewed as an $R'$-scheme) is given by the composition $\mathsf{Alg}_{R'} \to \mathsf{Alg}_R \to \mathsf{Set}$ where the first functor is restricting scalars along $R \to R'$ and the second functor is $X$.
More concretely this means that $(X \otimes_R R')(A, \iota) = X(A, \iota \circ \varphi)$ for any $R'$-algebra $(A, \iota \colon R' \to A)$.
In your situation we have $R = R' = k(t)$ and the morphism $\varphi \colon R \to R'$ is given by $t \mapsto t^m$ (just as you write).
This means that $G'(k(t), \mathrm{id}_{k(t)}) = G(k(t), \varphi)$.
Now we also consider the ring $k(t^{1/m})$ that we view as a $k(t)$-algebra via the natural inclusion $\gamma \colon k(t) \to k(t^{1/m})$.
Then the morphism $k(t) \to k(t^{1/m})$ gives by $t \mapsto t^{1/m}$ gives an isomorphism of $k(t)$-algebras $(k(t), \varphi) \to (k(t^{1/m}), \gamma)$ and thus induces an isomorphism of groups $G(k(t), \varphi) \to G(k(t^{1/m}), \gamma)$.
I think the confusing thing about all this is that when we base change along an endomorphism of a ring $R$ we suddenly need to very carefully remember that an $R$-algebra really consists of a ring together with a morphism from $R$ and that one ring $A$ can very well carry multiple $R$-algebra structures.
