As stated in the title, I am wondering the main difference between Beilinson conjecture and eTNC. If I read correctly, I can see that there are many literature treating both conjectures in the same line  special values of Lfunctions. If my impression is completely wrong, I am sorry for this stupid question. But, if they have some commonness, it would be greatly helpful if anyone can clarify their difference.

3$\begingroup$ I think it would be good to be more precise and, for instance, give references to the statments of the conjectures. Probably you mean Beilinson's conjecture (see Nekovar's survey) which says something about the rationality of the quotient of Lvalues by periods and regulator. The etnc is much finer giving integrality statements and an arithmetic interpretation of the leading terms in terms of Tamagawa numbers and it allows for the action of a possibility noncommutative group on everything. $\endgroup$– Chris WuthrichCommented Dec 24, 2017 at 11:11
1 Answer
The key difference between these conjectures is the coefficient ring that is involved. You also are leaving out an important "middle" conjecture  the (nonequivariant) Tamagawa number conjecture, as formulated in Bloch and Kato's article in the Grothendieck Festschrift  and knowing what this conjecture says might clarify the relation between the other two conjectures a bit.
In Beilinson's conjecture, the coefficient ring is $\mathbf{Q}$: the conjecture predicts the leading terms of $L$functions up to rational factors, in terms of the determinant of a regulator matrix defined using a $\mathbf{Q}$basis of some motivic cohomology space.
In the BlochKato Tamagawa number conjecture (TNC), the coefficient ring is $\mathbf{Z}_\ell$ for some prime $\ell$, and the conjecture predicts the leading terms of $L$functions up to $\ell$adic units. The extra integral structure comes from comparing motivic cohomology with Galois cohomology, which (unlike motivic cohomology) works well with $\mathbf{Z}_\ell$ coefficients.
In the equivariant Tamagawa number conjecture (ETNC), one introduces further extra structure by considering a Galois extension $K / k$ and studying the arithmetic of a motive over $k$ baseextended to $K$, which means that all the cohomology groups are modules over the group ring $\mathbf{Z}_\ell[G]$ where $G = \operatorname{Gal}(K/k)$ is some finite group. Then the conjecture relates the leading terms of equivariant (i.e. groupringvalued) $L$functions to the isomorphism classes of cohomology groups as $\mathbf{Z}_\ell[G]$modules.
(You could also formulate an "equivariant Beilinson conjecture" involving $\mathbf{Q}[G]$modules, but it would be fairly trivally equivalent to the original Beilinson conjecture for each of the motives $M \otimes \rho$ where $\rho$ varies over representations of $G$. For the same reason, ETNC reduces to TNC when $\ell \nmid \#G$. The interesting cases are when $G$ is an $\ell$group, in which case there are nontrivial congruences between $\mathbf{Z}_\ell$representations of $G$ which the ETNC detects.)
So, to sum up, the triple of conjectures (Beilinson conj, TNC, ETNC) all seek to relate $L$values to arithmeticallyinteresting cohomology modules, but the cohomology modules are over the rings $(\mathbf{Q},\ \mathbf{Z}_\ell,\ \mathbf{Z}_\ell[G])$ respectively.

1$\begingroup$ @david Of course I agree with almost all of this nice answer, but I think that the ETNC even for $\ell\nmid \# G$ says something a bit stronger than TNC for the twists. For instance for two representations of a finite $G$ that are conjugate over $\overline\mathbb{Q}$, then ETNC with coefficients in $\mathbb{Q}[G]$ will also say that the order of vanishing of the two twisted $L$functions are equal and that the leading terms divided by periods and regulators are conjugate. I don't think that would follow from the TNC, would it? $\endgroup$ Commented Dec 28, 2017 at 14:36

$\begingroup$ I think this is part of the TNC, because Galoisconjugate Artin representations will correspond to the same Artin motive (the coefficient field of a motive is an "abstract" number field, it doesn't come with a distinguished embedding into $\mathbf{C}$). $\endgroup$ Commented Jan 1, 2018 at 20:35