suppose $k$ is a finite field, and $\widetilde{k}$ a finite extension of $k$ (say, of degree $n$). How does the set of $k$-rational points of $\mathrm{Spec}(\widetilde{k})$ look like ? (As a $k$-rational point corresponds to a morphism $\mathrm{Spec}(k) \mapsto \mathrm{Spec}(\widetilde{k})$, it also corresponds to a $k$-morphism $\widetilde{k} \mapsto k$, but field morphisms are always injective, right ? Or should one see the latter as a $k$-algebra morphism ?)
Thanks !
Questions like this are often implicitly meant in the context of schemes over $\operatorname{Spec(k)}$.
So, since both schemes are affine, this question is about $k$-algebra homomorphisms $\bar{k} \to k$ (where both rings are $k$-algebras in the suggested way). There are none. (unless $n=1$, in which case there is just the one)
And, for the sake of precision, the above is meant with the convention that $k$-algebra homomorphisms are also ring homomorphisms — that is, they preserve $1$.
But if you really did mean general schemes, a.k.a. schemes over $\operatorname{Spec} \mathbb{Z}$, $\operatorname{Spec(\bar{k})}$ can have $k$-points — that is, there can be ring homomoprihms $\bar{k} \to k$.
For example, if $k = F(x^2)$ and $\bar{k} = F(x)$ (with the field extension given by inclusion), there is a homomorphism $\bar{k} \to k$ sending $x \mapsto x^2$.