Let me first describe how I got to that problem. We know that Majorana Lagrangian (here I choose left-handed but for right-handed problem is analogue) $${\cal L}=\psi_{L}^{\dagger}i\bar{\sigma}^{\mu}\partial_{\mu}\psi_{L}-\frac{m}{2}\left[\psi_{L}^{T}\epsilon\psi_{L}+\psi_{L}^{\dagger}\epsilon\psi_{L}^{*}\right]\tag{1}$$ (where $\epsilon =\scriptscriptstyle\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}$) doesn't obey $U(1)$ symmetry, therefore Majorana Fermion doesn't cary a charge understood in a sense of electric like charge.
Nevertheless, it obeys $SO(2)$ symmetry (acting on $\psi$ by 2x2 representation of rotation), thus by the Noether theorem, there should be conserved current and charge of that symmetry. This is how I calculate it:
Noether current is given by the formula $$j^{\mu}=\frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\psi_{L,\alpha})}\epsilon_{\alpha\beta}\psi_{L,\beta}-K^{\mu},\tag{2}$$ where $K^{\mu}$ is $0$ as given symmetry is not only a symmetry of action but also the symmetry of Lagrangian and $\epsilon$ is an infinitesimal form of a rotation. Now we could be tricky. In the case of Majorana fermions, $\psi$ and $\psi^{\dagger}$ are not independent (relation given by the Majorana equation) therefore we could differentiate over $\psi^{\dagger}$ to calculate the current. Alternatively, we could move the derivative to $\psi^{\dagger}$ by integration by parts of the kinetic term of our Lagrangian.
All in all, we got the Noether current equal to $0$ and therefore conserved charge equal to $0$. My first question is did I calculate that charge correctly? And then if yes, what does it mean that a Noether current and charge are equal to 0? In my opinion, it doesn't give us any information about the dynamics of a system. Does it give any information about that symmetry? What does it mean in that particular example? And finally, are we able to predict that a given symmetry gives $0$ charge?