You are right that additional laws are necessary. This is because, as you point out, from the perspective of Noether's theorem, rotational symmetry is something that is distinct and irreducible in its own right to translational symmetry. A simple counterexample to the irreducibility of rotational symmetry is "taxicab geometry", which is a space in which instead of the distance formula being

$$d(P, Q) = \sqrt{(Q_x - P_x)^2 + (Q_y - P_y)^2}$$

it is

$$d_T(P, Q) := |Q_x - P_x| + |Q_y - P_y|$$

where $P = (P_x, P_y)$ and $Q = (Q_x, Q_y)$ are whatever two points we want to consider. The latter has only a finite rotational symmetry group, but its translational symmetry is as good as Euclidean geometry.

However, three laws are not required, though it is definitely more intuitive and natural-feeling to start with such a presentation. One extra law is sufficient:

• The forces exerted by two bodies upon each other act only along the line between them. [1]

That is to say, if $\mathbf{F}_{12}$ is the force that body 1 exerts on body 2, that, using the cross product to check parallelism,

$$\mathbf{F}_{12} \times \mathbf{r}_{12} = \mathbf{0}$$

which you can see is literally the statement that there is no torque ($\mathbf{r} \times \mathbf{F})$ in the system resulting from the two bodies alone, i.e. there are no self-torquing systems. (You don't need a corresponding statement for $\mathbf{F}_{21}$ because Newton's third law already constraints that from $\mathbf{F}_{12}$).

You are right that additional laws are necessary. This is because, as you point out, from the perspective of Noether's theorem, rotational symmetry is something that is distinct and irreducible in its own right to translational symmetry. A simple counterexample to the reducibility of rotational symmetry is "taxicab geometry", which is a space in which instead of the distance formula being

$$d(P, Q) = \sqrt{(Q_x - P_x)^2 + (Q_y - P_y)^2}$$

it is

$$d_T(P, Q) := |Q_x - P_x| + |Q_y - P_y|$$

where $P = (P_x, P_y)$ and $Q = (Q_x, Q_y)$ are whatever two points we want to consider. The latter has only a finite rotational symmetry group, but its translational symmetry is as good as Euclidean geometry.

However, three laws are not required, though it is definitely more intuitive and natural-feeling to start with such a presentation. One extra law is sufficient:

• The forces exerted by two bodies upon each other act only along the line between them. [1]

That is to say, if $\mathbf{F}_{12}$ is the force that body 1 exerts on body 2, that, using the cross product to check parallelism,

$$\mathbf{F}_{12} \times \mathbf{r}_{12} = \mathbf{0}$$

which you can see is literally the statement that there is no torque ($\mathbf{r} \times \mathbf{F})$ in the system resulting from the two bodies alone, i.e. there are no self-torquing systems. (You don't need a corresponding statement for $\mathbf{F}_{21}$ because Newton's third law already constraints that from $\mathbf{F}_{12}$).