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A projective transformation is a morphism of $P^n$ to $P^n$, for some $n$, determined by an $(n + 1) \times (n + 1)$ invertible matrix $A$ in the obvious way. The sets $Q$, $R$ are projectively equivalent if and only if there exists a projective transformation $f$ such that $f(Q) = R$.

I would like to know useful criteria to determine if two projective varieties are isomorphic or not.

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Your last sentence is inconsistent with the title of the question; I guess "isomorphic" should be replaced by "projectively equivalent". –  Artie Prendergast-Smith Jul 8 '13 at 16:53
You might look at Wilcysnki's book Projective Differential Geometry of Curves and Ruled Surfaces to get a sense of how complicated the projective invariants of curves can get. But Wilcynski is only thinking about local geometry of $C^{\infty}$ immersed curves in real projective space, so a very different story globally. –  Ben McKay Apr 20 at 11:03

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Equivalence of projective varieties does imply that the varieties are isomorphic. But the converse does not hold. For example $\mathbb{P}^1$ is isomorphic to the conic defined by $(xy - z^2)$ in $\mathbb{P}^2$ but there is clearly no way that these are projectively equivalent because the degrees do not match. I'm not sure if that answers your question completely but I think it might be helpful to people who stumble onto this page.

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