The following question is related to this post Showing an isomorphism between exterior products of a free module about the existence of a canonical isomorphism of the exterior product of free modules and in particular details the pathological behavior of certain isomorphisms given conditions on the rank of the module.
Let $F$ be a free $R$-module of rank $ n\geq 3$ where $R$ is a commutative ring with identity.
Let $T(F) = \oplus_{k=0}^{\infty} T^{k}(F)$ where $T^k(F) = F \otimes F \otimes \ldots \otimes F$ is tensor product of $k$ modules.
Let $\wedge F $ denote the exterior algebra of the $R$-module $F$, that is the quotient of the tensor algebra $T(F)$ by the ideal $A(F)$ generated by all $x \otimes x$ for $x \in F$.
Suppose $1\leq p \leq n-1$ and that $n \neq 2p$
How do we show there cannot exist a bijective module homomorphism $f: \wedge^p F \rightarrow \wedge^{n-p} F$ such that $f \circ (\wedge^p \psi) = (\wedge^{n-p} \psi ) \circ f $ for all $\psi \in Aut(F)$?