Is there an "easy" way to count points on a twists of elliptic curves? Say we consider the curve $\newcommand{\F}{\mathbb F} E(\F_p):\ y^2=x^3+x$ and her twist $E'(\F_{p^2})=x^3+2^{1/4}x$. (Some may see, this is a curve from the KSS-16 family.) Our basic-setting is: We know p, $\#E$ and the trace of frobenius $t$.
Let me understand first some different things. If I take the definition of a degree of a twist (attached), as I know it, this is a quartic/degree four twist? The fourth root of 2 has degree 4.
Def: Twist of Degree {2,4,6}
Consider $y^2=x^3+ax+b$ then the twists are given by $y^2 = x^3+a_ix+b_i$
If $ab\neq 0$, then $a_i=a, b_i=b/c^i$ where $deg(c)=2$ and $i=1,2$
If $b=0$, then $a_i=a/c^i$, where $deg(c)=4$ and $i=1,...,4$
If $a=0$, then $b_i=b/c^i$, where $deg(c)=6$ and i=1,...,6$
If $E$ is an elliptic curve over $\Bbb F_q$ with $p+1-t$ points, its quadratic twist has $p+1+t$ points.
Quartic twists only occur in curves with CM over $\Bbb Z[i]$. This multiplies the eigenvalues of Frobenius by powers of $i$. So if the trace of Frobenius is $t$, then $t=2a$ where $a\pm ib$ are eigenvalues of Frobenius. One of twisted curves has eigenvalues $b\pm ia$ and so has $p+1-2b$ points. The other has $p+1+2b$ points.
Cubic and sextic twists only occur over curves with CM over $\Bbb Z[\omega]$, $\omega=\frac12(-1+i\sqrt3)$. Similar considerations apply as in the quartic case, but multiplying by sixth roots of unity this time. I'll omit the details...