If $X$ is a projective scheme over a field $k$ (which we may assume is algebraically closed), then under an embedding $i: X \hookrightarrow \mathbb{P}^n_k$, we may write $X = Proj(R/I)$ where $R = k[x_0, x_1, \ldots, x_n]$ and $I$ is the homogeneous defining ideal.

Suppose we consider a second embedding $j: X \hookrightarrow \mathbb{P}^n_k$, corresponding to a second presentation $X = Proj(R/J)$ with $J$ another homogeneous ideal of $R$. (I am assuming that $n$ and hence $R$ are the same as before, i.e., I am only comparing embeddings into the same ambient projective space.)

Can anything algebraically precise be said about the relationship between $I$ and $J$?

If the closed subschemes $Proj(R/I)$ and $Proj(R/J)$ of $Proj(R)$ are the same, then it is well-known (Hartshorne ex. II.5.10) that $I$ and $J$ have the same saturation, so *saturated* homogeneous ideals correspond 1-1 with closed subschemes of $Proj(R) = \mathbb{P}^n_k$. Thus the fact that $I$ and $J$ define the same closed subscheme is equivalent to a purely commutative-algebraic fact about $I$ and $J$. If we no longer assume that the subschemes are the same, but just that both arise from embeddings of the same projective scheme $X$ into $\mathbb{P}^n_k$, is there also an equivalent commutative-algebraic relationship between $I$ and $J$?

If we knew that the ideal sheaves $\tilde{I}$ and $\tilde{J}$ on $Proj(R)$ were isomorphic, then Hartshorne ex. II.5.9(b) would imply that the graded ideals $I$ and $J$ were "asymptotically" isomorphic, in the sense that there would exist an integer $d_0$ such that the graded pieces $I_d$ and $J_d$ would be isomorphic for all $d \geq d_0$ (my only promising lead so far). But there seems to me to be no reason for $\tilde{I} \simeq \tilde{J}$ to hold, since changing the embedding will change the ideal sheaf.