Are photon energies conserved in general relativity?

As I understand it, both Maxwell's wave equation and the null geodesics of general relativity are scale invariant.

Thus an electromagnetic wave can be shifted along a null geodesic without changing the laws of physics that it obeys.

Does this imply that the energy/momentum of photons is conserved as they travel through spacetime relative to the co-ordinate system of some observer A?

The phenomenon of gravitational redshift could be due to the difference between an observer's local energy, as demonstrated by the frequency of his clocks, and the constant energy of the photon, both described relative to observer A's co-ordinates.

Answer

In a curved spacetime, the wave four-vector $k^a$ for a solution to Maxwell's equation (in the appropriate geometric-optics limit) satisfies the geodesic equation:

$$(k^a \nabla_a )k^b = 0.$$ Effectively, this means that "the derivative of $k^a$ in the direction $k^a$ is zero". (Note that the first term is just a directional derivative in the direction of $k^a$. What's more, it is always possible to set down a set of coordinates in a "tube" surrounding the worldline of this geodesic, called Fermi normal coordinates, such that all the Christoffel symbols vanish at points along the geodesic. (See here for a reference to how this is done for null geodesics.) In this case, if $\lambda$ is our coordinate along the null geodesic, then the geodesic equation as expressed in these coordinates is just $$\frac{d k^\mu}{d\lambda} = 0.$$ In other words, the components of the wave vector are constant in these coordinates. So in this sense, the energy & momentum of the photon is conserved in the "photon frame" (whatever that means.)

But of course, we're not photons, and so we observe gravitational redshift. In a lot of cases, though, we have a stationary spacetime, in which (roughly speaking) there's a time coordinate $t$ that the metric doesn't depend on. Formally speaking, we want there to be a vector field $t^a$, pointing in the direction of increasing $t$, such that $\nabla_a t_b + \nabla_b t_a = 0$. In this case, $t^a$ is called a Killing vector field, and it's not too hard to show in such an instance that

$$k_a t^a = \text{const.}$$ In this sense, a quantity related to the energy of the photon is conserved as the photon travels. For example, if we're just dealing with flat spacetime, then $t^\mu = (1,0,0,0)$ and we have $k^t = \text{const.}$, i.e., the energy of photons is constant for observers whose four-velocity points in the $t$-direction. More generally, in a static spacetime, we'll have something like $k_t = k^t g_{tt} = \text{const.}$ instead. Similarly, if we have a spacelike Killing vector field (for example, $x^a = (0,1,0,0)$ in flat spacetime), then we can define a conservation law for particles that travel along geodesics which reduces to conservation of momentum in the case of flat spacetime.

Finally, it's important to note that the actual energies & momenta that an observer will measure are related to the observer's four-velocity $u^a$ and their set of spatial basis vectors. For example, we have $\omega = u^a k_a$. Since $u^a u^b g_{ab} = -1$ by definition, this means that in general an observer whose four-velocity is in the $t$-direction will have $u^\mu = ((-g_{tt})^{-1/2},0,0,0)$, and such an observer will then see the photon as having

$$\omega = k_t u^t = \frac{\text{const.}}{\sqrt{-g_{tt}}}.$$ In this interpretation, the gravitational redshift is in fact solely due to the clocks of various observers "running slow" due to the $tt$-component of the metric being different. However, it's important to note that the conserved quantity $k_t$ is not necessarily an energy as seen by any particular observer (even the observer who emitted it in the first place.)

A good reference for this business with Killing vectors and conserved quantities is Hartle's Gravity: an Introduction to Einstein's General Relativity. Killing vectors are explained in Section 8.2.