I just organized my comments into an answer/review (though the existing answers are already perfect).

**Schemes over $k$.**

The category $Sch_S$ of schemes over a given scheme $S$ is by definition the comma category over $S$, i.e. objects are morphisms of schemes $\pi:X\to S$ and arrows are morphisms of schemes making the resulting diagram commute.
For $S=\mathrm{Spec}k$ we get the category $Sch_k$ of $k$-schemes. We will denote by $Sch$ the category of plain schemes over the integers (though in real life I'd not dare touching anything that is not, say, over $\mathbb C$...).

There is a forgetful functor $u:Sch_k \to Sch$, $(X\to \mathrm{Spec}k)\mapsto X$ sending a $k$-scheme to its underlying ($\mathbb Z$-) scheme. Notice that while e.g. $Spec\mathbb{C}\in Sch_\mathbb{C}$ is a one point scheme, $u(Spec\mathbb{C})\in Sch$ is a pretty messy object.

There is also a base change functor $b_k:Sch\to Sch_k$, $X\mapsto (X\times_{Spec \mathbb Z}Spec k\to Spec k)$. Notice that $b_k$ is definitely not the inverse of $u$ (e.g. $\mathbb{C}\otimes_{\mathbb Z}\mathbb{C}\neq \mathbb C$, and $\mathbb{Z}\otimes_{\mathbb Z}\mathbb C\neq \mathbb Z$).

Given $X$ over $k$, there are the two corresponding functors of points, $h_{X/k}:Sch_k^{op}\to Set$ and $h_{X/\mathbb Z}:Sch^{op}\to Set$. The former is not "the restriction" of the latter: $Sch_k$ is not a subcategory of $Sch$, rather a comma category.

Notice that $h_{X/k}\neq h_{X/\mathbb Z}\circ u$ (where I've still called $u:Sch_k^{op}\to Sch^{op}$): take e.g. $X=Speck$ and evaluate both functors at $T=Speck$. What you can say in general is $h_{X/k}\subseteq h_{X/\mathbb Z}\circ u$ is a subfunctor. Likewise, $h_{X/\mathbb Z}\neq h_{X/k}\circ b_k$: take e.g. $X=Speck$ and evaluate at $T=Spec\mathbb Z$.

What's the relationship between $\mathbb{P}^n_{\mathbb Z}$ and $\mathbb{P}^n_k$? The latter is just the image of the former under the functor $b_k$. But the functor of points of $\mathbb{P}^n_k$ is not a restriction of the functor of points of $\mathbb{P}^n_{\mathbb Z}$, it's really another functor: it gives the set of only morphisms *over* $k$ as opposed to all the morphisms.

Notice that the automorphism group of a $k$-scheme $X$ *as a $k$-scheme* i.e. in $Sch_k$ may be completely different from the automorphism group of $X$ as a plain scheme. Consider for example $X=Spec\mathbb{C}$: $Aut_{\mathbb C}(X)$ is trivial, while $Aut_{\mathbb Z}(X)$ is all the ring automorphisms of the $\mathbb Z$-algebra $\mathbb C$.

The same happens for projective space. Consider the $k$-scheme $\mathbb{P}^n_k$. If we consider it in $Sch_k$, then its $Aut$ is $PGL_n(k)$. I don't know what happens over the integers, but if we consider $\mathbb{P}^n_k$ as a scheme over the prime field $k_0$ of $k$, then $Aut$ is a semidirect product of $PGL_n(k)$ and the Galois group of $k$ over $k_0$. Apparently, the $Aut_{k_0}$ is the automorphism group of the abstract projective geometry (in the sense of incidence structures) induced by $\mathbb{P}^n_k$.

**Automorphisms functor.**

This is an aspect orthogonal to the above one. Let's consider everything in $Sch_k$ for simplicity.

The automorphisms functor of $X\in Sch_k$ is defined by $\mathbf{Aut}_k(X):S\mapsto Aut_S(X\times_k S)$, where $Aut_S$ denotes automorphisms in $Sch_S$ for every $S\in Sch_k$.

Now you see that, whether this is representable or not, we can talk about the $k$-points of $\mathbf{Aut}_k(X)$ and it is obvious from the definition that $(\mathbf{Aut}_k(X))(k)=Aut_k(X)$ as abstract groups. So, the automorphisms functor is designed to have as its $k$-points precisely the automorphisms of $X$ as an object of $Sch_k$.

What about more general points of $\mathbf{Aut}_k(X)$? You can think of $X\times_k S$ as a trivial $X$-bundle over $S$. An $S$-point $\sigma$ of $\mathbf{Aut}_k(X)$ is an isomorphism $\sigma:X\times S\to X\times S$ that commutes with projections to the base $S$. So you can see $\sigma$ as a "bundle automorphism" of $X\times_k S$, or as a family of automorphisms of $X$ parametrized by $S$.