I'm not an algebraic geometer so let's call $X$ a real manifold. I don't think that really matters it could be a topological space or even a set for what I am about to say. Assume that $E \to X$ is a rank $n$ real vector bundle. For any $x \in X$ let $E_x$ be the fibre of $E$ over $x$. The natural structure you have in this situation is that if $f \colon E_{x} \to E_{y} $ and $g \colon E_{y} \to E_{z}$ are isomorphisms then you can compose to get $g \circ f \colon E_{x} \to E_{z}$ also an isomorphism. From this follows the fact that you have a groupoid whose objects are all $x \in X$ and whose morphisms from $x$ to $y$ are $Isom(E_x, E_y)$ (or $Isom(E_y, E_x)$ depending on how you like to compose morphisms.)
(1) $Isom(E, E)$ is a perfectly reasonable object, it's a bundle of groups over $X$. But it doesn't capture all the information such as isomorphisms from $E_x$ to $E_y$ where $x \neq y$.
(2) $Isom(p_1^*E, p_2^*E) $ is the union over all $x, y \in X$ of $Isom(E_x, E_y)$ and if $f \in Isom(E_x, E_y)$ then the two maps are $p_1(x, y) = f \mapsto x$ and $p_2(x, y) = f \mapsto y)$or the other way around depending on how you define the projections.
(3) I am not sure of the answer to this but it seems reasonable to me that this groupoid captures information about the symmetries of $E \to X$.
I don't think that $Isom(p_1^*E, p_2^*E) $ being a principal $GL(n, \mathbb{R})$ bundle is correct. I don't see any reason why $Isom(E_x, E_y)$, for example, is acted on by $GL(n, \mathbb{R})$.
