Let $C$ be a hyperelliptic curve $y^2 = f(x)
$ defined over $\mathbb{Q}$, where $f(x) \in \mathbb{Q} [x]$ is a polynomial of degree $n=5$ or $6$, and $J = Jac(C)$ its Jacobian.
I know Zarhin's result [Hyperelliptic Jacobians without complex multiplication], which states that if the Galois group $G= Gal(f)$ of $f$ is either the Symmetric group $S\_n$ or the alternating group $A\_n$, then the endomorphism ring of $J$ is $\mathbb{Z}$ (that is, $J$ has no complex multiplication).


My question is that whether there is a condition on $G$ under which $J$ has CM.


The reason why I ask this is the follwing;
In [Wamelen, Examples of genus two CM curves defined over the rationals], Wamelen found 19 curve (of gunus 2) whose Jacobian has CM.
In all of these examples, the Galois group of $f$ is either the cyclic group $C\_4$ of order $4$ or the Frobenius group $F\_5$ of order $20$.
So my question is that;


Is there any example of $C$ without CM and with $G = C\_4$ ? 



Hyperelliptic Jacobians with (or without) CM
The curve $$y^2 = x^5 + 2x^4 + 2x^3 +2x^2 + 2x +1$$ has $G= C\_4$ and its Jacobian $J$ has endomorphism algebra ${\rm End}^0(J) \cong \mathbb{Q}$.


For the curve $$y^2 = x^5 + x^4 + x^3 + x^2 + x$$ we have $G= C\_4$ and its Jacobian $J$ has endomorphism algebra ${\rm End}^0(J) \cong \mathbb{Q} \times \mathbb{Q}$.


As explained in Jedrzej's answer, it is possible to impose conditions on $G$ which restrict the possible endomorphism algebras. A relevant example here is: if $C:y^2=f(x)$ is a hyperelliptic curve with $f(x) \in \mathbb{Q}[x]$ of degree 5 or 6, and $|G|$ is odd, then the Jacobian of $C$ does not have CM. In fact, a stronger statement is true: if $f(x)$ has degree 5 or 6 with coefficients in some number field $K$, Galois group $G$ of odd order, and $J$ has CM, then $K$ must contain a real quadratic field.


