Automorphisms of Schemes and their $A$-points At first, I figured that an automorphism of a scheme $X$ would be a homeomorphism $f:|X| \to |X|$ of topological and an isomorphism of sheaves $f^{\#}: \mathcal{O}_X \to f_*(\mathcal{O}_X)$. 
However, if $X= \mathbb{P}_k^n$, then the automorphisms of $X$ are actually defined as follows: 
Let $\textbf{Aut}(\mathbb{P}^n)$ denote the functor taking a scheme $S=\operatorname{Spec}A$. where $A$ is a commutative algebra over an algebraically closed field $k$, to the group of automorphisms $\operatorname{Aut}_A(\mathbb{P}_A^n)$ of $\mathbb{P}_A^n$ over $A$.
I think the definition means that we need to look at the "automorphisms of all $A$-points of $\mathbb{P}_k^n$". I'm not sure if this last phrase is correct, I've just heard it being said. 
However, how does this naturally follow from the "naive" interpretation of what an automorphism of a scheme $X$ would be? I can't figure out why or when the $A$-points of a scheme are of significance. 
For example, when Hartshorne discusses the automorphism group of $\mathbb{P}_k^n$ he determines it to be $PGL(n, k)$, i.e he really only considers the $k$-points of the group scheme $\textbf{PGL}(n)$. 
 A: The object you've defined is not the group of automorphisms of $\mathbb{P}^n$; among other things, it is a group-valued functor, not a group. Here is a simpler example of this sort of thing:
In any category $C$, if $X, Y$ are two objects you can consider the set $\text{Hom}(X, Y)$ of morphisms $X \to Y$. If $C$ has finite products, then one can furthermore consider the presheaf sending an object $Z$ to the set $\text{Hom}(X \times Z, Y)$, which one can equivalently think of as $\text{Hom}_Z(X \times Z, Y \times Z)$, where $\text{Hom}_Z$ means the hom is taken in the slice category over $Z$. If this presheaf is representable, its representing object is called an exponential object $Y^X$. The exponential object is strictly more information than just the homset, which one recovers by taking global points $\text{Hom}(1, Y^X) \cong \text{Hom}(X, Y)$.
Similarly, in any category $C$, if $X$ is an object one can consider the group $\text{Aut}(X)$ of automorphisms of $X$, and if $C$ has finite products, then one can consider the group-valued presheaf sending an object $Z$ to the group $\text{Aut}_Z(X \times Z)$. If this presheaf is representable, we might call its representing object an "automorphism object" of $X$. Again it contains strictly more information than just the automorphism group of $X$, which again one recovers by taking global points. 
$\mathbb{P}^n$ as a $k$-scheme has $PGL_n(k)$ as its group of automorphisms, and if I'm not mistaken its automorphism object furthermore exists and is the group scheme $PGL_n$ over $k$. Similarly one can construct the group scheme $GL_n$ over $k$ by considering automorphisms of base changes of $k^n$ as a $k$-vector space object in affine schemes over $k$. 
A: 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 $S$.
A: I am not sure I understand your problem, but maybe what you are missing is that for a projective variety $X$, the functor $A\mapsto \operatorname{Aut}(X_A) $ is representable - that is, there is a group scheme $\operatorname{\underline{Aut}}_X$ such that $\operatorname{\underline{Aut}}_X(A)=\operatorname{Aut}(X_A) $ (this follows directly from the theory of the Hilbert scheme). In particular, the group of $k$-points $\operatorname{\underline{Aut}}_X(k)$ is the usual automorphism group $\operatorname{Aut}(X) $.
