Automorphism of $\mathbb{P}_A^n$ Let $k$ be a field. 
The proof that $Aut(\mathbb{P}_k^n)=PGL(n+1:k)$ relies on the fact for any automorphism $\alpha$ of $\mathbb{P}_k^n$, $\alpha^*(\mathcal{O}_{\mathbb{P}_k^n}(1)) = \mathcal{O}_{\mathbb{P}_k^n}(1)$. 
It is not necessarily true that $\pi^*O_{\mathbb{P}_A}(1)\simeq O_{\mathbb{P}_A}(1)$ if $\pi:\mathbb{P}_A^{n}\to\mathbb{P}_A^{n}$ is an automorphism. 
See counterexample: https://math.stackexchange.com/questions/979958/automorphism-of-the-projective-space-mathbbp-an
In particular, we can't use the same type of proof to show that $Aut(\mathbb{P}_A^n)=PGL(n+1:A)$. Furthermore, I don't think such an equality will exist for an arbitrary ring $A$. 
What is the intuitive reason as to why we might not have such equality for arbitrary rings? What are the conditions on A for which this might be true? (other than the obvious one such as A a UFD or a field)
 A: Let me give an example of a Dedekind domain $A$ such that the map 
$$H^1(\operatorname{Spec}A, \mathbf{G}_m) \to H^1(\operatorname{Spec} A, \text{GL}_2)$$
is not injective. In particular, this will give a counterexample to the statement that the points of $\text{PGL}_2$ valued in a ring $A$ coincide with ${\rm{GL}}_2(A)/A^{\times}$ (an equality we should expect to fail sometimes when ${\rm{Spec}}(A)$ has non-trivial line bundles for exactly the same reason that in the setting of smooth manifolds it isn't generally true for a smooth manifold $M$ with non-trivial line bundles that every smooth map $M \to \mathbf{RP}^n$ lifts to a smooth map to $\mathbf{R}^{n+1}-\{0\}$).
Recall that the map of sheaves $\mathbf{G}_m$ to $\text{GL}_2$ is given by the diagonal embedding. Therefore, identifying $H^1(\operatorname{Spec}A, \mathbf{G}_m)$ with the class group of $A$,
the map  $$H^1(\operatorname{Spec}A, \mathbf{G}_m) \to H^1(\operatorname{Spec} A, \text{GL}_2)$$
is given as follows. Take an ideal $I$ and send it to the $\text{GL}_2$-torsor over $\operatorname{Spec} A$ associated to $I \oplus I$. What is this $\text{GL}_2$-torsor? Well, it is simply $\operatorname{Spec} \operatorname{Sym} (I \oplus I)$. So now here comes the punchline. Let me further assume that $A$ is a Dedekind domain with a $2$-torsion ideal in its class group, i.e. an ideal $I$ such that $I^2$ is principal. Then in this case $I \oplus I \cong A^{\oplus 2}$ as $A$-modules. Therefore  the ideal $I$ is mapped to $$\operatorname{Spec}( \operatorname{Sym} (A\oplus A)) \cong \mathbf{A}^2_A$$ which is  trivial in $H^1(\operatorname{Spec} A, \text{GL}_2).$ Thus, when the class group of $A$ has non-trivial 2-torsion (as happens for rings of integers with even class number) then injectivity fails.
A: Let me summarize the comments of R. van Dobben de Bruyn. The $A$-automorphisms  of $\mathbb{P}^n_A$ correspond in a one-to-one way to couples $(u,L)$, where $L$ is an element of $\operatorname{Pic}(A) $ and $u:A^{n+1}\rightarrow L^{n+1}$ an isomorphism. In other words, there is an exact sequence
$$1\rightarrow \operatorname{GL}(n+1,A)/A^*\rightarrow  \operatorname{Aut}(\mathbb{P}^n_A)\xrightarrow{\ \lambda \ } \operatorname{Pic}(A)\, ,  $$where the image of $\lambda $ is the subgroup of elements $L\in\operatorname{Pic}(A) $ such that $L^{n+1}\cong A^{n+1}$ (note that this implies in particular $L^{{\scriptscriptstyle\otimes }(n+1)}\cong A$).
