My answer is that the state of the system remains stable: that is, the ice remains ice and the water remains water.
Let's try to bring a few concepts together: since the whole system is at T=0 and isolated from the environment, a phase transition (in either direction) can't complete. Hence, everything moving away from the original state is driven by stochastic fluctuations. We want to prove that the original state is an equilibrium point: every fluctuation can produce a local phase transition, but this can't extend any further; actually it will disappear in a short time and the system will go back to the original state.
Let's assume that a stochastic fluctuation brings a localised infinitesimal volume of the water below T=0, lets say that the local temperature is $T_l=-\epsilon$. (Everything of course will work the same way around in the ice, melting locally inside the water at $T_l=+\epsilon$).
Normally a metastable phase is present and the fluctuation will average out to zero before anything the intermolecular bonds can be destroyed or created and the phase transition doesn't even start. But let's suppose that this is not the case and that a local phase transition actually starts happening. Then we will have a locally $T_l=-\epsilon$ and a ice sphere of ice of radius $R=\delta$ with both $\epsilon$ and $\delta$ much smaller than the typical dimensions of the system.
The phase transition follows the theory of nucleation (see Wikipedia). The results of this theory tell us that exists a critical radius $R_c$ which has the following property:
- if $R<R_c$ the ice sphere will disappear and its radius shining exponentially in time
- il$R>R_c$ the ice sphere will instead grow exponentially and the phase transition will that place
again the second case can't happen: because if the radius starts to grow, it will soon encounter the border $\Sigma$ of the infinitesimal volume of the fluctuation and the phase transition will stop.
A different (but connected) example
The following is not directly connected to the answer, but should remark that the described situation is stable.
Suppose that we have the entire water at a state $T_c=-\epsilon$. Let's try to calculate the mean time to have a phase transition - that is, a fluctuation which creates an ice sphere of radius $R>R_c$.
The difference in free energy is of the form:
$$\Delta F = T_s R^2 -\Delta f R^3$$
where $T_s$ is the surface tension coefficient (the partial ordering of the molecules of ice against the water phase cost a certain amount of free energy which scales with the surface, hence the $R^2$ dependence) and $\Delta f$ is the difference between free energy per unit volume of the ice and the water (see this image).
The $R_c$ is defined the maximum point of the function $\Delta F(R)$ because for $R>R_c$ the derivative is negative and an increase of the radius of the ice sphere diminish the free energy of the system and for $R<R_c$ the opposite holds.
From this definition ve get that the critical radius is:
$$R_c= \frac{2 T_s}{3\Delta f}$$
Hence we get
$$\Delta F_c=\Delta F(R_c)\sim\frac{1}{\Delta f^2}$$
which is the variation of the free energy given by the fluctuation needed to generate the phase transition. Near $T=0$ we can assume $\Delta f \sim \Delta T=\epsilon$ (see the same picture for clarification).
Arrhenius law tells us that the average waiting time for a fluctuation is:
$$\tau= \tau_0 e^{\beta\Delta F}$$
hence:
$$\tau \sim e^{\frac{\beta}{\epsilon^2}}$$
This tells us that is the temperature variation is small we should wait an insanely long time to see the transition happening.
I want to remark that this is not a proof, but we can convince ourselves that if a system with the entire water undergoing the fluctuation has a characteristic waiting time that long, the system described in the question is stable.
Hope that helps!
