What is the difference between symmetric states and geometrically uniform states?

On the one hand, in this article, symmetric states are defined as:

The $$M$$ symmetric quantum states which should be distinguished by means of the quantum measurement are described by statistical operators, \begin{align*} \hat{\rho}_j&=\left|\psi_j\middle\rangle\!\middle\langle\psi_j\right|\\ &= \hat{V}^{j-1}|\psi\rangle\!\langle\psi|\hat{V}^{\dagger j-1} \end{align*} where $$j=1,2,\ldots,M$$.

[...] The operator $$\hat{V}$$ is unitary and satisfies the relation $$\hat{V}^M=\hat{I}$$ where $$\hat{I}$$ stands for an identity operator.

On the other hand, in this article, geometrically uniform states are defined as:

We define a geometrically uniform (GU) state set as a collection of vectors $$S = \left\{|\phi_i\rangle = U_i|\phi\rangle,U_i\in\mathcal{G}\right\}$$, where $$\mathcal{G}$$ is a finite abelian (commutative) group of $$m$$ unitary matrices $$U_i$$, and $$|\psi\rangle$$ is an arbitrary state.

What is the difference between these two notions?

Note that $$\left\{\hat{V}^{j-1}\middle|j=1,2,\ldots,M\right\}$$ is a finite abelian group of $$M$$ unitary matrices. As such, the set $$\left\{\hat{V}^{j-1}|\phi\rangle\middle|j=1,2,\ldots,M\right\}$$ satisfies the definition of geometrically uniform states. This set is (taking bras instead of brakets), according to the definition, a set of symmetric quantum states.
However, it is not immediate to claim that the converse is true. It is quite easy to find a group $$\mathcal{G}$$ of unitaries that is both abelian and finite and that is not generated by a single element of this group. For instance, one might take $$U$$ to be a controlled-$$X$$ gate with the control qubit being the first one and the target qubit to be the second one and $$V$$ to be $$I\otimes X$$, that is an $$X$$ gate on the second qubit. You can check that $$\mathcal{G}=\{I, U, V, UV\}$$ is a finite abelian group and it's definitely not generated by a single element.
However, we're interested in applying these elements to a starting point $$|\psi\rangle$$, and it may just be that there is a unitary $$R$$ that generates the exact same states. For instance, with the example above, if we start from $$|\psi\rangle=\frac{|00\rangle+|11\rangle}{\sqrt{2}}$$, then $$R=\begin{pmatrix}0 & 0 & 0 & 1\\0 & 0 & 1 & 0\\1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\\end{pmatrix}$$ generates the same states.