Suppose we have two curves $C/\mathbb{Q}$ and $C'/\mathbb{Q}$ which are twists of each other i.e. they are isomorphic over a field extension $K/\mathbb{Q}$.
Suppose that $C$ has good reduction at a prime $l$. Is it true that $C'$ has good reduction at $l$ if and only if $l$ is unramified in $K$? It seems to me that this should be the case (for example by looking at quadratic twists of elliptic curves) but I am not sure why it is true. Could someone provide a proof or a reference for a proof?
Let $B$ be a Dedekind scheme with function field $K$. (Think of $B=\mathrm{Spec } \ \mathbb{Z}_{p}$ for simplicity, so that $K=\mathbb{Q}_p$.) Let $C$ be a smooth proper geometrically connected curve over $K$ of genus at least one with good reduction over $B$. Let $\mathcal{C}\to B$ be its (unique) smooth proper model.
Let $C'$ be a twist of $C$ over $K$. Let $L/K$ be a finite field extension such that $C'_L \cong C_L$ over $L$. One can use the theory of Neron models for curves due to Liu-Tong to prove the following (see https://arxiv.org/abs/1312.4822).
Lemma. Assume that the normalization $B'$ of $B$ in $L$ is (finite flat) etale over $B$. Then $C'$ has a smooth proper model over $B$.
Proof. Let $\mathcal{C}'\to B$ be the Neron model of $C'$ over $B$. Since $B'\to B$ is finite etale, the basechange $\mathcal{C}'\times_B B'$ is the Neron model of $C'_L$ over $B'$. Since $C'_L\cong C_L$, we have that $C'_L$ has good reduction over $B'$. In particular, its Neron model is proper over $B'$. Thus, $\mathcal{C}' \times_B B'\to B'$ is proper (as it is the Neron model of $C'_L$ over $B'$). We conclude that $\mathcal{C}'\to B$ was already proper. In particular, $\mathcal{C}'\to B$ is a smooth proper model for $C'$ over $B$. QED
Concretely: when $K=\mathbb{Q}_p$ and $B=\mathrm{Spec} \ \mathbb{Z}_p$, if $C'$ is a twist of $C$ which has good reduction after passing to an unramified extension of $\mathbb{Q}_p$, then it must have already had good reduction over $\mathbb{Q}_p$.
Consider the case of an elliptic curve $A$ over $\mathbb{Q}$ and a quadratic extension $K$ of $\mathbb{Q}$. Let $A_{\ast}$ be the Weil restriction of $A_{K}$. Then $c(A_{\ast})=N_{K/\mathbb{Q}}(c(A_{K}))d_{K/\mathbb{Q}}^{2}$. Here $c$ is the conductor and $d$ the discriminant. As $A_{\ast}$ is isogenous to the product of $A$ and its twist, and isogenies don't change conductors, I think this gives what your want. Reference: Milne, Invent. Math. 1972, Pptn.1.
The formula, in fact, holds for abelian varieties, so if a curve and its quadratic twist both have good reduction, then the quadratic field $K$ must be unramified.
