It's a (relatively) well-known fact that, if $a_0,\ldots,a_n\in\mathbb{N}$ share a common factor $d\in\mathbb{N}$ then $$\operatorname{Proj}k_a[x_0,\ldots,x_n] \cong \operatorname{Proj}k_{a/d}[x_0,\ldots,x_n]$$ where $k_a[x_0,\ldots,x_n]$ represents the polynomial ring with $\deg x_i=a_i$, and $k_{a/d}[x_0,\ldots,x_n]$ is where $\deg x_i=a_i/d$.
Now $k_{a/d}[x_0,\ldots,x_n]=k_a[x_0,\ldots,x_n]^{(d)}$ is the $d$-th truncation of $k_a[x_0,\ldots,x_n]$ where we use the convention that we divide the grading by $d$ in the truncation. There is a clear graded ring homomorphism $R\to R^{(d)}$ given by $x_i\mapsto x_i^d$, but what is an inverse to this?
Or is it in fact the case that these two rings are not isomorphic as graded rings, but we still have an isomorphism of their $\operatorname{Proj}$? This question seems to be asking the same thing but where we use the convention that the grading stays the same in the truncation, and I also don't have a copy of the Eisenbud on me, so I can't look up the reference given in the comments.
It seems like this question should be useful, which says that a graded ring homomorphism only has to be an isomorphism of graded rings for large enough grading, but I still can't see how to use this.