AFAIK, such a reaction has no name. However, if one were so inclined, one could call it a replacement reaction, because the $\ce{MgCO3}$ is being converted into $\ce{Mg(OH)2}$.
To calculate the "equilibrium point," which I take to mean the concentration of all the ions at equilibrium, we would need to know the $k_\text{sp}$ of both $\ce{MgCO3}$ and $\ce{Mg(OH)2}$, as well as the $\ce{pH}$ of the solution (which gives us $[\ce{OH^-}]$).
The solubility products of $\ce{MgCO3}$ and $\ce{Mg(OH)2}$ are $10^{-7.8}$, which is approximately $1.58 \cdot 10^{-8}$, and $5.61\cdot 10^{-12}$, respectively.
To calculate how much magnesium hydroxide precipitates, we would first need to find the concentrations of $[\ce{Mg^{2+}}]$ and $[\ce{CO3^{2-}}]$ using the $k_\text{sp}$ of magnesium carbonate. This gives us the solubility of $\ce{MgCO3}$, which is $1.26 \cdot 10^{-4} \,\, \ce{mol lit^{-1}}$. That means that we have $[\ce{Mg^{2+}}] = 1.26 \cdot 10^{-4} \,\, \ce{mol lit^{-1}}$ in the beaker.
Now, to find out how much $\ce{Mg(OH)2}$ will precipitate, we need to find the reaction quotient $Q_c$ of the reaction.
$$Q_c = [\ce{Mg^{2+}}][\ce{OH^-}]^2$$
(I cannot calculate $Q_c$ here because it depends on $[\ce{OH^-}]$, which will have to be given.)
If $Q_c > k_\text{sp}$, then the reaction
$$\ce{Mg(OH)_2(s) <=> Mg^{+2}(aq) + 2OH^-(aq)}$$
will proceed in the backward direction and magnesium hydroxide will precipitate out. This will continue until the concentration of magnesium ions decreases to such a value that $Q_c = k_\text{sp}$, at which point the reaction will be at equilibrium. Here, I've assumed that the concentration is too large to be appreciably changed by it precipitating out of the solution. The exact value of $[\ce{Mg^{2+}}]$ as well as how many moles of $\ce{Mg(OH)2}$ will precipitate can be calculated with this assumption. (However, if we want, we can do it without this assumption too. The calculations would become more messy and involve solving a quadratic.)
