Let $S$ be a scheme, let $\mathbb{P}_{S}^{n}$ be projective $n$-space over $S$, and let $s_{1},s_{2} : S \to \mathbb{P}_{S}^{n}$ be two sections of the projection $\mathbb{P}_{S}^{n} \to S$. What is an example of such $S,s_{1},s_{2}$ such that there does not exist $\varphi \in \mathrm{Aut}_{S}(\mathbb{P}_{S}^{n})$ with $\varphi s_{2} = s_{1}$?
Remark: Such $\varphi$ exists Zariski-locally on $S$, since such $\varphi$ always exists if $S = \operatorname{Spec} A$ where $A$ is a local ring. Indeed, since line bundles over local rings are trivial, each section $s_{\ell}$ (for $\ell = 1,2$) corresponds to an ordered $(n+1)$-tuple $a^{\ell} = (a_{0}^{\ell},\dotsc,a_{n}^{\ell})$ of elements in $A$ such that at least one $a_{i}^{\ell}$ is a unit of $A$. By using the unit $a_{i}^{\ell}$ to "perform row operations", there exists an invertible matrix $M \in \mathrm{GL}_{n+1}(A)$ such that $M a^{1} = a^{2}$. Then the image of $M$ under $\mathrm{GL}_{n+1}(A) \to \mathrm{Aut}_{A}(\mathbb{P}_{A}^{n})$ is the desired $\varphi$.
You have more or less said when this should be true.
For an example, take $S$ to be the spec of $A=\mathbb{R}[x,y.z]/x^2+y^2+z^2=1$, the co-ordinate ring of the real sphere. A section $s$ as above with $n=2$ corresponds to a surjection $s:A^3\to A$ and two sections $s_1, s_2$ have such a $\phi$ is equivalent to an automorphism $\phi:A^3\to A^3$ such that $s_2\phi=s_1$. This would imply that the kernels of $s_1,s_2$ are isomorphic. But it is well known that the two maps $s_1=(1,0,0)$ and $s_2=(x,y,z)$ have non-isomorphic kernels, since one of them is free and the other is the tangent bundle to the sphere, which is not free.
This is an addendum for the question below. If $\phi$ is an automorphism of $X\times S$ to itself, where $X$ is irreducible and projective and $S$ is affine, then $\phi$ induces an automorphism of $S$. If this induced automorphism is identity, then it is an anutomorphism over $S$. So, in the above situation, all automorphims in your question is given by $GL_3(A)$. I hope this answers your question.