Noether's theorem under arbitrary coordinate transformation Noether's theorem states that every differentiable symmetry of the action of a physical system has a corresponding conservation law.
Suppose our action is of the form 
$S = \int d^4x\, \mathcal{L}(\phi,\partial_\mu\phi).\tag{1}$
if $x \rightarrow x'$ then if $S \rightarrow S'$ where
$S' = \int d^4x'\, \mathcal{L'}(\phi',\partial_\mu\phi').\tag{2}$
But from calculus we know that $S=S'$ so does that mean that every change of variable correspond to a conserved quantity? why the quantities conserved under Poincare transformation, for example, is more especial?
 A: The action shown in the question is a functional of $\phi$, not of $x$. A change of the integration variable $x$ is just a relabeling of the index set. It does not transform the dynamic variables $\phi$ at all, so no: a change of variable does not correspond to a conserved quantity.
More explicitly, if $y(x)$ is a monotonic smooth function of $x$, then
$$
 \int d^4y\ {\cal L}\left(\phi\big(y(x)\big),\, 
  \frac{\partial}{\partial y_\mu}\phi\big(y(x)\big)\right)
 =
 \int d^4x\ {\cal L}\left(\phi(x),\frac{\partial}{\partial x_\mu}\phi(x)\right)
\tag{1}
$$
identically, for any ${\cal L}$ whatsoever (as long as it depends on $x$ only via $\phi$). This is just a change of variable (a relabeling of the index-set), and there is no associated conserved quantity.
In contrast, suppose that the action has this property:
$$
 \int d^4x\ {\cal L}\left(\phi\big(y(x)\big),\, 
  \frac{\partial}{\partial x_\mu}\phi\big(y(x)\big)\right)
 =
 \int d^4x\ {\cal L}\left(\phi(x),\frac{\partial}{\partial x_\mu}\phi(x)\right).
\tag{2}
$$
Unlike equation (1), equation (2) is not identically true for any ${\cal L}$ and any $y(x)$, though it may be true for some choices of ${\cal L}$ and $y(x)$. The transformation represented in equation (2) replaces the original function $x$, namely $\phi(x)$, with a new function of $x$, namely $\phi\big(y(x)\big)$.
This is the kind of transformation we have in mind when we talk about Poincaré invariance and its associated conserved quantities: it is a change of the function $\phi$ which we then insert into the original action, not a change of the integration variable.
A: *

*Note that in the context of Noether's theorem that some quasisymmetry transformations are so-called trivial quasisymmetry transformations, and corresponds to a trivial conservation law, cf. e.g. my Phys.SE answer here. 

*In particular, if we consider a passive coordinate transformation of the system, then the action $S$ is trivially invariant, and we can e.g. use the trick with $x$-dependent infinitesimal  $\epsilon(x)$ to conclude that the corresponding full Noether current $J^{\mu}$ vanishes identically.
