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)



Automorphism of $\mathbb{P}_A^n$
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 $L^{{\scriptscriptstyle\otimes }(n+1)}\cong A$).


