What is the correspondence between the conserved canonical energy-momentum tensor, which is $$ T^{\mu\nu}_{can} := \sum_{i=1}^N\frac{\delta\mathcal{L}_{Matter}}{\delta(\partial_\mu f_i)}\partial^\nu f_i - \eta^{\mu\nu}\mathcal{L}$$ (the four conserved Noether currents corresponding to four possible spacetime translations)
where $\{f_i\}_{i=1}^N$ are the $N$ matter fields in the theory, and we assume $f_i\mapsto\alpha^\nu\partial_\nu f_i$ for translations,
and stress-energy tensor from the Einstein-Hilbert action, which is: $$ T_{\mu\nu}\equiv-\frac{2}{\sqrt{-g}}\frac{\delta\mathcal{L}_{Matter}}{\delta g^{\mu\nu}} $$
In particular, how do you get that the two are equal (are they?) for Minkowski space, for which there is no variation in the metric?
Answer
You should think at the way the Noether current is obtained. When an infinitesimal symmetry transformation is made spacetime dependent, that is the parameters $\omega^a$ that control the symmetry are taken as functions of the spacetime point $\omega^a=\omega^a(x)$, the action is not longer left invariant $$ \delta S=-\int d^D x\, J^{a\,\mu}(x)\partial_\mu \omega^a(x) $$ but rather it provides the definition of the current $J^{a\,\mu}$ that is conserved on-shell.
Now, let's look at the case of the energy momentum tensor: in this case, the translations $x^\nu\rightarrow x^\nu+\omega^\nu$ are made local $x^\nu\rightarrow x^\nu+\omega^\nu(x)$ so that $$ \delta S=-\int d^D x\, T_\nu^\mu(x)\,\partial_\mu \omega^\nu(x)\,. $$ Actually, one looks for a symmetric tensor $T^{\mu\nu}=T^{\nu\mu}$ so that one can rewrite the expression above in the following form $$ \delta S=-\frac{1}{2}\int d^D x\, T^{\nu\mu}(x)\,(\partial_\mu \omega_\nu(x)+\partial_\nu\omega_\mu)\,. $$ Now, here is the catch: if we were to transform the spacetime metric $g_{\mu\nu}$ (equal to $\eta_{\mu\nu}$ in the case at hand) as if $x^\nu\rightarrow x^\nu+\omega^\nu(x)$ was just an infinitesimal change of coordinates, that is $$g_{\mu\nu}\rightarrow g_{\mu\nu}-(\partial_\mu \omega_\nu(x)+\partial_\nu\omega_\mu)\,,$$ then the action (rendered coordinates independent by the inclusion of the metric in the usual way such as $d^D x\rightarrow d^D x \sqrt{|g|}\,,\ldots$) would be left invariant $$ \delta S=-\frac{1}{2}\int d^D x\, (\partial_\mu \omega_\nu(x)+\partial_\nu\omega_\mu)\left. \left(\sqrt{|g|}\,T^{\mu\nu}(x)+2\frac{\delta S(x)}{\delta g_{\mu\nu}}\right)\right|_{g_{\mu\nu}=\eta_{\mu\nu}}=0\,. $$ From this equation it follows that the current associated with spacetime translations can be written as $$ T^{\mu\nu}=\left. -\frac{2}{\sqrt{|g|}}\frac{\delta S}{\delta g_{\mu\nu}}\right|_{g_{\mu\nu}=\eta_{\mu\nu}} \qquad \mbox{evaluated on the bkg }g_{\mu\nu}=\eta_{\mu\nu}\,. $$ It should be apparent that this definition gives a symmetric energy-momentum tensor that matches the one appearing the Einstein's equations. From the derivation above it should also be clear that the alternate versions of $T_{\mu\nu}$ arise because the definition of $T^{\mu\nu}$ via the variation of the action when the translation is made spacetime dependent does not uniquely fix it. For example, given a valid $T_{\mu\nu}$, one can always define another one $T^{\mu\nu}\rightarrow T^{\mu\nu}_B=T^{\mu\nu}+\partial_\rho B^{\rho\mu\nu} $ with an arbitrary $B^{\rho\mu\nu}=-B^{\mu\rho\sigma}$ which also gives $$ \delta S=-\int d^D x\, T^{\mu\nu}(x)\partial_\mu\omega_\nu(x)=-\int d^D x\, T_B^{\mu\nu}(x)\partial_\mu\omega_\nu(x) $$ up to an integration by parts. Einstein's equations break this degeneracy and nicely identify ``the'' energy-momentum tensor.
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