Let $Y$ be a scheme. For a locally free $\mathcal{O}_Y$-module $\mathcal{E}$ of rank $n$, the scheme $\mathbb{V}(\mathcal{E})$ is defined as $\mathbb{V}(\mathcal{E}) := \mathbf{Spec}(S^\bullet\mathcal{E})$.
Now if $f: X \rightarrow Y$ is a morphism of schemes, we can consider the locally free $\mathcal{O}_X$-module $f^*\mathcal{E}$, and the associated scheme $\mathbb{V}(f^*\mathcal{E})$.
Are $f^*$ and $\mathbb{V}$ compatible in the sense that there exists a (unique?) morphism $\mathbb{V}(f): \mathbb{V}(f^*\mathcal{E}) \rightarrow \mathbb{V}(\mathcal{E})$ that commutes with the canonical maps $p_X:\mathbb{V}(f^*\mathcal{E}) \rightarrow X$ and $p_Y:\mathbb{V}(\mathcal{E}) \rightarrow Y$? I.e. does $$p_Y\circ\mathbb{V}(f) = f \circ p_X$$ hold?
My problem is that I do not know how to relate the sheaves $f^*\mathcal{E}$ and $\mathcal{E}$, as both live on different spaces, let alone the schemes defined by those.
Such a morphism is not unique. For instance, if $X = Y = Spec(k)$ and $f$ is the identity, then $V(E) = V(f^*E)$ is just an affine space, and any its automorphism satisfies the required compatibility.
On the other hand, there is a canonical choice of a morphism induced by the natural isomorphism of sheaves of algebras $$ f^*(S^\bullet E) \cong S^\bullet(f^* E). $$