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 $G = C_4$ ?