I can't understand the comment on page 409, Gravitation, by Misner, Thorne, Wheeler
It follows that the ten components $G_{\alpha\beta} =8\pi T_{\alpha\beta}$ of the field equation must not determine completely and uniquely all ten components $g_{\mu\nu}$ of the metric.
On the countrary, $G_{\alpha\beta} =8\pi T_{\alpha\beta}$ must place only six independent constraints on the ten $g_{\mu\nu}(\mathcal{P})$, leaving four arbitrary functions to be adjusted by man's specialization of the four coordinate functions $x^{\alpha}(\mathcal{P})$.
I can't understand it. I think we can always solve the field equation with appropriate initial/boundary conditions to get unique $g_{\mu\nu}$. After all those are just second order differential equations. To be specific, let me try to construct a counter example, the vacuum Einstein equation,
$$G_{\mu\nu}=0$$
If we apply the initial conditions $g_{\mu\nu}|_{t=0}=\eta_{\mu\nu}$ and ${\dot{g}_{\mu\nu}}|_{t=0} =0$, obviously the flat spacetime $g_{\mu\nu}=\eta_{\mu\nu}$ should be the solution. If the solution $g_{\mu\nu}$ is unique, what's the alternative solution?
If there does exist an alternative solution, does it come from "specialization of the four coordinate functions"?
Update:
user23660 constructed an explicit alternative solution, which is
$$
g_{00}=(f'(t))^2,\quad g_{ij}=-\delta_{ij}
$$
with other components being zero.
The function $f$ only need to satisfy $f'(0)=1,f''(0)=0$, that makes this metric compatible with the initial data; other than that, it's completely arbitrary! And we see that it does come from the coordinate transformation $t=f(\tau)$
To get the solution to be $\eta_{\mu\nu}$, we need to put further constraints on the metric directly in this coordinate system, like $g_{00}=1,g_{0i}=0$.
This redundant degrees of freedom(gauge) result from the contracted Bianchi identity, as explained in the following paragraph in MTW page 409,
$$G^{\alpha\beta}{}_{;\beta}=0$$
is true automatically, and so the equation of motion of the matter fields $T^{\alpha\beta}{}_{;\beta}=0$ doesn't really put restrictions on the evolution of the metric. Therefore, there are only 6 independent equations!
Best Answer
Of course, the metric $\eta_{\mu\nu}$ is not a unique solution for Einstein vacuum equations compatible with your given initial data. And yes, we can interpret the alternatives as arising from coordinate functions.
Let us take the simplest of such function: redefine time by introducing new 'time' variable $\tau$ through a relation $t=f(\tau)$ (spacial coordinates we will keep intact). The metric in new coordinates $(\tau,x,y,z)$ would be $$ ds^2=(f'(\tau))^2 d\tau^2 - \delta_{ij}dx^i dx^j. $$ It is, obviously, a different metric. And by choosing the function $f$ satisfying some simple conditions ($f(0)=0$, $f'(0)=1$, $f''(0)=0$) this metric will be compatible with your initial data.
But at the same time it is equally obvious that this metric still corresponds to the same space-time - the Minkowski space-time (at least locally).
Addition. To make a solution of Einstein equations unique one can use coordinate conditions (which are analogous to gauge fixing conditions in EM theory). These work as constraints on metric imposed in addition to Einstein equations.
Also, if you are interested in initial data - time evolution formulation of general relativity, I recommend looking at the ADM formalism.