Scaling of null vectors By definition any vector with $g_{\mu\nu}n^\mu n^\nu = 0$ is a null vector.
For simplicity let's look at flat spacetime. Is there any physical difference between two null vectors like these two?
\begin{align}
n^\mu = (1,1,0,0), \qquad m^\mu = (2,2,0,0).
\end{align}
Both are null vectors which could be used to describe a photon moving in the x-direction.
My gut feeling tells me that only $n^\mu$ should be a proper null vector which can be used for a photon four-velocity. If I were to use $m^\mu$ to define the photon 4-momentum, $p^\mu = \nu m^\mu$, I would get different results for physical quantities, e.g. energy, as if i were to use $n^\mu$.
Is there some other constraint that I am missing? Especially when trying to define a null vector in a generic spacetime $g_{\mu\nu}$ it is not clear to me which of the infinite many choices is correct. Again for a photon moving towards the x-direction both of the following 4-velocities would be valid,
\begin{align}
n^\mu = (1,n^1(g_{\mu\nu}),0,0), \qquad m^\mu = (n^0(g_{\mu\nu}),1,0,0).
\end{align}
 A: A photon should travel along null geodesic. Let's say that $\textbf{n}$ is a null vector field and let $u$ be a scalar field which parameterizes each of it's integral curve $\mathcal{C}$. Then $\mathcal{C}$ is a null geodesic, if tangent vector $n^a$ is parallelly propagated along this trajectory:
$$n^a\nabla_a n^b = \kappa n^b$$ $\kappa$ is the "surface gravity". The parameter $u$ is affine if $n^a\nabla_a n^b=0$. One way to achieve this is by rescaling the null vector $n \to n'=\alpha n$, which leads to $$n'^a\nabla_a n'^b=\alpha^2 (\kappa + n^c\nabla_c\ln\alpha)n^b$$We can choose our scalar field $\alpha$ such that RHS of the above equation vanishes. So our photon momentum should correspond to the rescaled null vector $\textbf{n}'=\alpha\textbf{n}$ (unique upto an overall constant scaling factor)
Given that $n'$ satisfies geodesic equation, we can think of it as a generator for a Null Hypersurface $\mathcal{H}$. In 2+2 formalism, it is customary to consider a normalization defined by $n'^a=\frac{dx^a}{dt}$, or equivalently, $\langle \textbf{d}t,\textbf{n}'\rangle =1$, where $t$ is observer's time coordinate (See section 4.2 of https://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.603.9113&rep=rep1&type=pdf). This condition will take care of the  overall constant rescaling factor.
A: The two null vectors
\begin{align}
n^\mu = (1,1,0,0), \qquad m^\mu = (2,2,0,0).
\end{align}
might represent two different photon 4-momenta:

*

*$n^\mu = (1,1,0,0)$ has energy $n^\mu t_\mu=1$ (using +---) and x-momentum $1$

*$m^\mu = (2,2,0,0)$ has energy $2$ and x-momentum $2$
$m^\mu$ is akin to violet light and $n^\mu$ is akin to red light.
So, $m^\mu$ has a higher frequency, in this frame [with 4-velocity $t^\mu$] and in all frames.
That is a physical difference.
These two null vectors are proportional, and related by a boost.... 
in particular, they are related by a Doppler factor.
The notion of "4-velocity" for a photon might need a definition.

*

*For a massive particle, the 4-velocity is the unit-vector along the particle's 4-momentum---that is, the 4-momentum divided by the invariant-mass.

*For a massless particle, the same construction doesn't work since the massless particle has zero invariant-mass. What does "unit" mean now, in an invariant way?

It's not clear what "proper null vector" means.
