Why is the FLWR metric dependent on the observer? The FLWR metric is:
$$ds^2=-dt^2+a(t)d\vec{x}^2$$
and has a function whose dependence is on $t$. Different observers measure the time elapsed differently. Does that mean each observer measures the expansion of the universe differently?
 A: That's for comoving observers, if they also have peculiar velocity relative to the CMB background their clocks will be off, see here, and also the direction of motion will be lorentzcontracted by the Gammafactor relative to comoving observers. The comoving clocks have the longest proper time possible, clocks with peculiar velocity have less.
A: The time coordinate chosen for the FLRW metric is comoving i.e. it is such that you have the isotropy of space made manifest due to an observer moving along with the Hubble flow. A non-comoving observer would not be able to write down the spacetime metric as such in their own coordinates.
Since the expansion of space happens isotropically, a non-comoving observer would have to necessarily find a comoving coordinate basis first, and then calculate the rate of expansion. If they don't do that, in general they would measure different rates of expansion for each spatial direction of their coordinate basis.
So even though the scale factor $a$ is coordinate-dependent, the rate of expansion $H = \dot{a}/a$ is always defined for the comoving frame of reference.
A: FLRW universes have matter in them (usually). That matter has a particular state of motion, which breaks the symmetry of speeds.
This has nothing to do with general relativity as such. The same thing happens in a special-relativistic world that is full of matter. Wherever you go, you can measure your speed relative to the local matter, so velocities aren't equivalent. If the matter is expanding, rotating, or doing the Charleston, then the speed at which you'll see that happen depends on your motion. There is no symmetry forcing your observations to be independent of velocity.
In a FLRW universe that contains matter, the broken symmetry singles out a preferred $t$ coordinate, which is the one used in that metric. If there is no matter ($ρ=p=0$, but $Λ$ may still be nonzero), then all velocities are equivalent. The form of the FLRW metric may suggest otherwise, but actually the $t$ coordinate is meaningless without matter. You can put different FLRW coordinate systems with different $t$ coordinates on the same physical spacetime, just as you can put different inertial coordinate systems with different $t$ coordinates on a Minkowski vacuum, and none is more correct than any other.
