How does the Galois group act on morphisms of varieties? Let's say that I have two algebraic varieties $X$ and $Y$, both defined over a field $k$. Let $K/k$ be a Galois extension. How does the $\text{Gal}(K/k)$ act on morphisms $f:X_K \rightarrow Y_K$? It seems that if I have a morphism, I can look at the stabilizer of $f$ in $\text{Gal}(K/k)$ and then take the fixed field of that subgroup. If I call that fixed field $k'$, then is the morphism $f$ defined over $k'$?
When you are talking about morphisms defined over a smaller field, is it always with respect to the Galois action above? For example, if the Galois group acts differently on a morphism, and I do the same procedure with looking at the fixed field of the stabilizer, does that morphism descend?
 A: Let $X$ be scheme over a field $k$ and let $K$ be a Galois extension of $k$. One then can define a morphism for each $\sigma$ in $\mathrm{Gal}(K/k)$
$$\sigma_{X,K}:X_K\to X_K$$
as $\sigma_{X,K}:=\mathrm{id}_X\times \mathrm{Spec}(\sigma^{-1})$ acting on $X_K=X\times_{\mathrm{Spec}(k)}\mathrm{Spec}(K)$. One can then check that
$$\sigma\mapsto \sigma_{X,K}$$
defines a homomorphism $\mathrm{Gal}(K/k)\to \mathrm{Aut}(X_K)$ where $\mathrm{Aut}(X_K)$ is the group of automorphisms of $X_K$ as an abstract scheme.
Suppose now that $Y$ is another $k$-scheme. The action of $\mathrm{Gal}(K/k)$ on $\mathrm{Hom}_K(X_K,Y_K)$ is given by
$$\sigma f:=\sigma_{Y,K}\circ f\circ \sigma_{X,K}^{-1}$$
It is then true that, with some minor hypotheses perhaps, that the map
$$\mathrm{Hom}_k(X,Y)\to \mathrm{Hom}_K(X_K,Y_K)$$
is injective with image precisely $\mathrm{Hom}_K(X_K,Y_K)^{\mathrm{Gal}(K/k)}$. Most of this can be deduced from [1, §4.4].
This should be sufficient to answer the rest of your questions, I believe.
[1] Poonen, B., 2017. Rational points on varieties (Vol. 186). American Mathematical Soc..
