$\DeclareMathOperator\Norm{Norm}$Suppose $E/\mathbb{Q}(j(E))$ is a CM elliptic curve and $d$ is a non-square. Let $E\_d$ denote the twist of $E$ by $\mathbb{Q}(j(E))(\sqrt{d})$. I know if $d$ is relatively prime to the conductor of $E$, then we have $$N\_{E\_d} = d^2N\_{E}$$
However, by computational investigations, there appear to be certain primes that always divide the norm of the conductor regardless of twist.


As a computational example: I will take a curve with CM by $K=\mathbb{Q}(\sqrt{-5})$. I take the curve with $j$-invariant
$$j(E)= 5887918080 (31261995198\sqrt{5} - 69903946375)$$
(i just picked the one with smallest absolute norm). Taking the absolute norm of the conductor of $E$ gives the factorization
$$\Norm(N\_E)=5^2 \* 7^4 \* 11^4 \* 19^2 \* 41^2 \* 47^4 \* 199^2 $$
Taking $d=77$, I get the twisted curve $E\_d$ which has conductor
$$\Norm(N\_{E\_d}) =5^2 \* 19^2 \* 41^2 \* 47^4 \* 199^2 $$
No matter what integer I twist by, I cannot get rid of these factors. Do I need to twist by something in $\mathbb{Q}(j(E)) =\mathbb{Q}(\sqrt{5})$?


Other examples seem to indicate that there are primes of bad reduction common to all twists. For example for CM elliptic curves over $\mathbb{Q}$ with CM by $\mathbb{Q}(\sqrt{-q})$, where $q$ is an odd prime. $q$ is a prime of bad reduction among all curves with the same $j$-invariant.


My question:
What is known about the primes of bad reduction common to all CM curves with the same $j$-invariant?



Primes of bad reduction for CM elliptic curves
These are exactly the primes where the Neron model of $E$ has fiber type, in [Kodaira's table](https://en.wikipedia.org/wiki/Elliptic_surface), anything other than $I\_0$ and $I\_0^\*$. Here $I\_0$ represents good reduction and $I\_0^\*$ represents a quadratic twist of good reduction by a ramified extension (quadratic twisting by an unramified extension preserves the reduction type).


(At least, this is true away from primes above 2, which are surely more complicated).


The fiber type can be computed by Tate's algorithm, but to tell whether the type is $I\_0$ or $I\_0^\*$, you can simply perform a ramified quadratic twist and check whether either the curve or its quadratic twist has good reduction. The number you quadratic twist by doesn't matter, as long as it is $\pi$-adic valuation, where $\pi$ is the prime of bad reduction, is odd.


For $\pi$ in $\mathbb Q(\sqrt{-5})$ lying over a prime $p$ of $\mathbb Q$, unless $\pi$ is ramified (i.e. $\pi$ lies above $5$), then any number which has $p$-adic valuation odd will do the trick.


