Is the (isomorphism class of) a rational surface determined by its Picard group and its intersection form?

Question

As the title says: if I have two rational surfaces $X$ and $Y$ with $\mathrm{Pic}(X)\cong\mathrm{Pic}(Y)$ and such that the intersection forms on $X$ and $Y$ are equivalent, then does this force $X\cong Y$? I have tried looking at this for surfaces of Picard rank $2$ and $3$, and it seems to hold, but I can't seem to come up with a proof nor a counterexample.

Answer 1

Definitely not. For instance all smooth cubic surfaces have isomorphic Picard group but their isomorphism classes are parameterized by a 4-dimensional moduli space.