Let $Y$ be a vector field on a Riemannian manifold $(M, g)$. If $g(\nabla Y, \nabla Y)=0$, then $Y$ is covariantly constant, i.e. $\nabla Y=0$.
Now, let $V$ be a vector field on a Lorentzian manifold $(N, \gamma)$. We assume $V$ is a null vector. If $\gamma(\nabla V, \nabla V)=0$ where $\nabla$ is the covariant derivative of $\gamma$, what can we say about $V$?
Very little can be said about the vector field $V$. Below I am going to use index notation. In particular, nothing near to covariantly constant can be gained.
If you know $V$ is null, you have $$\tag{1}\label{eq:null} \gamma_{ab} V^a V^b = 0 $$ which implies $$\tag{2}\label{eq:1stder} \gamma_{ab} V^a \nabla_c V^b = 0 $$ On top of this you assumed that $$\tag{3}\label{eq:assm} \gamma_{ab} \gamma^{cd} \nabla_c V^a \nabla_d V^b = 0.$$
Consider now the vector field $$ W^a = f V^a $$ for some function $f$. Then $$ \nabla_c W^a = \nabla_c f V^a + f \nabla_c V^a. $$ So $$ \gamma_{ab} \gamma^{cd} \nabla_c W^a \nabla_d W^b = f^2 \gamma_{ab} \gamma^{cd}\nabla_c V^a \nabla_d V^b + 2 f \gamma_{ab} \gamma^{cd} \nabla_c f V^a \nabla_d V^b + \gamma_{ab} \gamma^{cd} \nabla_c f \nabla_d f V^a V^b $$ The three terms on the right vanishes by virtue of \eqref{eq:assm}, \eqref{eq:1stder}, and \eqref{eq:null} respectively. So you've found that
Proposition If $V$ is a null vector field that satisfies \eqref{eq:assm} then any scalar multiple of $V$ also satisfies \eqref{eq:assm}.
In fact, when $N$ is two-dimensional, suppose $V$ is any non-vanishing null vector field, then $V^\perp$ is equal to the span of $V$, so $\nabla_a V^b = \xi_a V^b$ for some one form $\xi$, which automatically implies $\gamma(\nabla V, \nabla V) = 0$. So when $N$ is two-dimensional, we have that any non-vanishing null vector field solves \eqref{eq:assm}.
The previous section showed that the "length" of such a vector field cannot be fixed, which is already incompatible with any rigidity condition like being covariantly constant. Next we show by explicit example that the direction of $V$ is also not fixed.
For simplicity let us fix $N = \mathbb{R}^{1,3}$ the standard Minkowski space with coordinates $(t,x,y,z)$ and $\gamma = - dt^2 + dx^2 + dy^2 + dz^2$. Consider an arbitrary smooth function $\phi:\mathbb{R}\to\mathbb{R}$, and set $$ V = \partial_t + \cos(\phi(t+z)) \partial_x + \sin(\phi(t+z)) \partial_y $$ which is clearly a null vector field. Then $$ \nabla V = \phi'(t+z) ( dt + dz) \otimes \big[ -\sin(\phi(t+z)) \partial_x + \cos(\phi(t+z)) \partial_y \big] $$ and as $(dt+dz)$ is a null one form, we have $\gamma(\nabla V, \nabla V) \equiv 0$.
