The Tamagawa numbers for an elliptic curve $E/\mathbb{Q}$ are given by $n_{p}(E) := [E(\mathbb{Q}_{p}):E^{0}(\mathbb{Q}_{p})]$, where $E^{0}(\mathbb{Q}_{p})$ is the group of points having good reduction. Let $\Delta$ be the discriminant of a given $E$. Let $E\otimes \chi_{D}$ be a quadratic twist of $E$ with discriminant $\Delta_{D}$.
Question: Is it true that for primes $p \nmid D$ we have $n_{p}(E\otimes\chi_{D}) = n_{p}(E)$?
Page 4 of this paper gives some insight on computing $n_{p}$. I know that $\Delta_{D} = \Delta D^{6}$ (I think?), and so if $p \nmid D$ then...? Perhaps knowing what happens to reduction types after twisting $E$ could help.
this overflow post could shed some light as well.