No. Work out the example of $G = E$ an elliptic curve/$k$. The exercise is a bit humorous. ^_•

Not all is lost though. A more general type of phenomena you may be interested in is that if you start with $T = $Coh($Y$) (coherent sheaves) than you can actually construct a stack $X$ which behaves as $X(S) =$ {fiber functors $T \to S$-mod}, where $S$ is a $k$-algebra.

If your scheme $Y$ was nice enough, then $X$ will actually be $Y$. (If I recall correctly, separatedness is critical.)

In fact this is what you are doing in the well known affine group reconstruction case as well: Under the hood you are using $BG$. This has only one point and the uniqueness of the fiber functor reflects this.

N̶o̶.̶ (YES, the answer above is correct.) The following example shows we cannot drop that condition: Work out the example of $G = E$ an elliptic curve/$k$. (The Rep category is severely degenerate in this case.)

[Edited. Some how at that late hour I completely missed the word 'affine' in the OP. Indeed, if the group is affine then his question is answered by the celebrated Tannaka theorems. I'll leave the info here though as it's nice to have examples of what fails.]

Not all is lost though. A more general type of phenomena you may be interested in is that if you start with $T = $Coh($Y$) (coherent sheaves) than you can actually construct a stack $X$ which behaves as $X(S) =$ {fiber functors $T \to S$-mod}, where $S$ is a $k$-algebra.

If your scheme $Y$ was nice enough, then $X$ will actually be $Y$. (If I recall correctly, separatedness is critical.)

In fact this is what you are doing in the well known affine group reconstruction case as well: Under the hood you are using $BG$. This has only one point and the uniqueness of the fiber functor reflects this.

No. Work out the example of $G = E$ an elliptic curve/$k$. The exercise is a bit humorous. ^_•

No. Work out the example of $G = E$ an elliptic curve/$k$. The exercise is a bit humorous. ^_•

Not all is lost though. A more general type of phenomena you may be interested in is that if you start with $T = $Coh($Y$) (coherent sheaves) than you can actually construct a stack $X$ which behaves as $X(S) =$ {fiber functors $T \to S$-mod}, where $S$ is a $k$-algebra.

If your scheme $Y$ was nice enough, then $X$ will actually be $Y$. (If I recall correctly, separatedness is critical.)

In fact this is what you are doing in the well known affine group reconstruction case as well: Under the hood you are using $BG$. This has only one point and the uniqueness of the fiber functor reflects this.