The lower-left corner is supposed to be a $1$, not a star (unless the central character is trivial, which it isn't in your case unless $\chi_0$ is quadratic).

Regardless, as LSpice notes, these representations do not have a conductor in that sense, unless the conductor of $\chi_0$ is $1$, because they are not invariant under any of these subgroups.

This is not so surprising as the theorem only says that generic representations have a conductor.

Maybe a better thing to try is to tensor your representation with the Steinberg representation, obtaining an infinitesimally equivalent representation which is generic, whose conductor is equal to $\min(c(\chi_0)^2, p)$.

The lower-right corner is supposed to be a $1$, not a star (unless the central character is trivial, which it isn't in your case unless $\chi_0$ is quadratic).

Regardless, as LSpice notes, these representations do not have a conductor in that sense, unless the conductor of $\chi_0$ is $1$, because they are not invariant under any of these subgroups.

This is not so surprising as the theorem only says that generic representations have a conductor.

Maybe a better thing to try is to tensor your representation with the Steinberg representation, obtaining an infinitesimally equivalent representation which is generic, whose conductor is equal to $\min(c(\chi_0)^2, p)$.