TITLE: Example of two subvarieties of $\mathbb{P}^2$ that are isomorphic but not projectively equivalent.
QUESTION [6 upvotes]: Two curves $C_1$, $C_2 \subset \mathbb{P}^2$ are called projectively equivalent if there is a projective change of coordinates $\phi: \mathbb{P}^2 \to \mathbb{P}^2$ so that $\phi(C_1) = C_2$. What is an example of $C_1$, $C_2 \subset \mathbb{P}^2$ so that $C_1$ is isomorphic to $C_2$ as a projective variety, but $C_1$ and $C_2$ are not projectively equivalent?

REPLY [5 votes]: Let $x$, $y$, $z$ be the homogeneous coordinates of $\mathbb{P}^2$. Consider the line $X_1 = \mathbb{V}(x)$ and the nonsingular quadric $X_2 = \mathbb{V}(xy - z^2)$. Both $X_1$ and $X_2$ are isomorphic to $\mathbb{P}^1$ (note that $X_2$ is the image of the second Veronese embedding of $\mathbb{P}^1$). However, since a linear change of coordinates on $\mathbb{P}^2$ must preserve the degrees of homogeneous equations, they are not projectively equivalent.