$\newcommand\im{\operatorname{im}}$Let $R = k[x_1,\ldots,x_n]$ as $\mathbf{Z}^n$-graded $k$-algebra, and $M$ be a finitely generated graded $R$-module. Then it is standard that there is a morphism $\phi\colon F \to G$ of finite rank free modules such that $\operatorname{coker} f \cong M$. After fixing bases for $F$ and $G$, we may represent $\phi$ by a matrix $\Phi$ with entries in $R$. As $M$ is $\mathbf{Z}^n$-graded, $\Phi$ will be a matrix of homogeneous entries. Concretely, if $F = \bigoplus_{j} R(f_j)$ and $G = \bigoplus_i R(g_i)$ for $f_j, g_i \in \mathbf{Z}^n$, then, using multi-index notation for exponents $\boldsymbol{x}^\alpha$, we get $\phi_{ij} \in k\boldsymbol{x}^{f_j-g_i}$ if $f_j-g_i \geq 0$, and $\phi_{ij} = 0$ otherwise. I can always find a minimal presentation, meaning that $\operatorname{im} f \subseteq (x_1,\dotsc,x_n)G$. Let's assume that $f$ is minimal.
However, $\Phi$ need not be unique; first, the morphism $f$ is not unique, and second, $F$ and $G$ can be subject to base change, which changes $\Phi$, but of course not $f$.
Question: Is there any way to define a canonical form of presentation matrix, in the sense that I can transform any presentation matrix into its canonical form, and two modules are isomorphic if and only if their canonical presentation matrices coincide. Of course, if $R = k[x]$, this is the case; this can be derived from existence of Smith Normal Form (see this question). However, for $R$ as above, no SNF need exist.
I can think of choosing an order on $\mathbf{Z}^n$ and performing base changes on $F$ and $G$ that eliminate as many entries from $\Phi$ as possible in a kind of algorithmic way; however, I am not sure how to see that the result satisfies the above criteria. Probably there is a Groebner-base view onto this (do submodules of free modules have unique reduced Groebner bases? Then I could probably work with a GB for $\operatorname{im} f$), but I know too little about Groebner bases.
Edit: I know that there is a unique reduced GB for $\im f$. However, different presentations $f'$ may have $\im f' \neq \im f$, only isomorphic. Therefore, I see no reason why $\im f$ and $\im f'$ should have the same reduced GB (I bet they don't).
