Let $R$ be a commutative ring, $M$ a free $R$-module of rank $n$ and $f \in \rm{End}(M)$. The adjugate $f^\sharp$ of $f$ is defined by the equalities $$ f^\sharp(x) \wedge y = x \wedge \Lambda^{n-1}f(y) $$ for all $x \in M$, $y \in \Lambda^{n-1}M$. From this definition (and by using the similar definition of the determinant of $f$) it is easy to show that $f^\sharp \circ f = \det(f) \cdot \rm{id}_M$. However, I am currently unable to show that also $f \circ f^\sharp = \det(f) \cdot \rm{id}_M$ holds, i.e. that $f$ commutes with $f^\sharp$.
I would be happy about any advice.
These type of questions are usually easier working over the `universal' ring. Consider $S=\mathbb{Z}[x_{ij}]$ for $1\leq i,j\leq n$ and a free module $F$ of rank $n$ over $S$. Then you have $g$ an endomorphism of $F$ defined by the matrix $(x_{ij})$ and you have a ring homomorphism $\phi:S\to R$ given by sending $x_{ij}$ to the $ij^{\mathrm{th}}$ entry of $f$. Now it is easy to see that what you need will follow if you can prove a similar statement for $g$.
As you said $g^{\#}\circ g=\det(g) \mathrm{id}_F$. Notice that $\det g,\det g^{\#}$ are non-zero and thus $g^{\#}$ is injective, since $S$ is an integral domain. Composing with $g^{\#}$ on the right, we have $g^{\#}\circ g\circ g^{\#}=\det (g)g^{\#}$, which yields $g^{\#}\circ(g\circ g^{\#}-\det (g) \mathrm{id}_F)=0$. Using injectivity of $g^{\#}$, one has $g\circ g^{\#}=\det(g) \mathrm{id}_F$.