Hyperelliptic Jacobians with (or without) CM

Example of a (hyper)elliptic curve with $G = C_4$ and no CM:

For

$C : y^2 = x^4 - x^3 + x^2 - x + 1$

one has $G = C_4$ (see here). On the other hand, $C$ is the elliptic curve with label 200.b2 and has no complex multiplication.

(I searched for a polynomial with Galois group $C_4$ using lmfdb.org, then computed the conductor and j-invariant of the corresponding curve in Magma and then searched for the curve in lmfdb)

Comment: I believe that the philosophy is that the bigger the Galois group is, the smaller is the ring of automorphisms. In other words, it is possible to impose a comdition on the Galois group such that the endomorphism ring will be small. It seems unlikely that any condition on the Galois group will give you a large endomorphism ring.

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.