How are FRW metric and Minkowski metric physically different? According to GR, matrices are coordinate invariant. Does this mean we can transform FRW metric to Minkowski metric with a coordinate transformation like 
$$dx'=dx\cdot a(t), dy' = dy\cdot a(t), dz' = dz\cdot a(t)$$ 
If yes, then why do we say that the two represent different spacetimes? If no, then why not? 
 A: To decide if two metrics are related by a change of frame and/or coordinate transformation is called the equivalence problem. It can be solved using the Cartan-Karlhede algorithm.
Given a metric $g$ expressed in some coordinates $x_i$, the algorithm computes a set of invariantly defined curvature invariants expressed as functions of $x_i$. For example, the scalar curvature $R = R(x_i)$. To decide if two metrics are equivalent, compute this set for both metrics and consider the set of equations \begin{align*}
R(x_i) & = R'(y_i) \\
\Psi_1(x_i) & = \Psi_1'(y_i) \\
& \vdots
\end{align*}
where primed quantities refer to the second metric, which is expressed in the coordinates $y_i$. (The full set of equations is typically much larger, but also typically many of the equations are $0 = 0$.)
If you can solve for the $y_i$ as functions of the $x_i$ or vice versa (or just show that a solution exists), you have established that the metrics are equivalent. If it is clear that a solution does not exist (for example, one of the equations could be $1 = 0$) the metrics are not equivalent.

For the particular case of the FLRW metrics compared to the Minkowski metric, one of the equations is $$0 = (k + \dot{a}^2)$$
where $k$ and $a$ are the quantities that appear in the FLRW line element
and another one is $$0 = k+\dot{a}^2 + a\ddot{a}$$
Combining these it must be that $a$ is a constant and $k = 0$. This corresponds to the Minkowski metric with the spatial part expressed in spherical coordinates with the radial coordinates being scaled a factor $a$ relative to each other.

This conclusion can also be reached from the Friedmann equations. For the Minkowski metric $\rho - \Lambda / \kappa = p + \Lambda / \kappa = 0$ and the Friedmann equations imply $k = 0$, $a$ constant.
A: It makes no sense to demand that $dx' = a(t)\ dx$. Suppose that your coordinate transformation was of the form $x' = x'(x)$. Then you would have $dx' = \frac{dx'}{dx}\ dx$, but $\frac{dx'}{dx}$ would have to be a function of $x$ only, and so it couldn't be $a(t)$. Now suppose we tried to fix that by doing a transformation $x'  = x'(x,t)$. Now $dx' = \frac{\partial x'}{\partial x}\ dx + \frac{\partial x'}{\partial t}\ dt$, and we get a $dt$ term which will mess things up. So your proposed coordinate transformation is not actually one.
The moral here is that defining a transformation by asking for a corresponding transformation of the differentials is only guaranteed to work when a single variable is involved each time. For example, if for some reason you wanted a new coordinate $x'$ such that $dx' = x^2\ dx$, then you could simply integrate to find $x' = \frac13 x^3$. But in your case, if you're going to demand that $dx' = f\ dx + g\ dt$ where both $f$ and $g$ depend on $(x,t)$ and (here) $g=0$, then this is only realizable if $\partial_x g = \partial_t f$, which is not true in your proposed transformation.
