Trying to check current conservation under symmetry transformation 
Consider a simple scalar field and its Lagrangian $L=\frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi$. Then say you have the following transformation
$$x^{\mu}\rightarrow e^{\omega}x^{\mu},\tag{1}$$
$$\phi\left(x\right)\rightarrow e^{-\omega}\phi\left(e^{\omega}x\right).\tag{2}$$
What is the associated conserved current?

Attempt:
I compute the Euler-Lagrange (EL) equations,
$$\partial_{\mu}\left(\frac{\partial L}{\partial\left(\partial_{\mu}\phi\right)}\right)=\frac{\partial L}{\partial\phi}\quad\Leftrightarrow\quad\partial_{\mu}\partial^{\mu}\phi=0,\tag{3}$$
write the transformations infinitesimal as
$${\phi\left(x\right)\rightarrow\left(1-\omega+\ldots\right)\phi=\phi-\omega\phi}\quad\Rightarrow\delta\phi=-\omega\phi\tag{4}$$
$${\delta L=0}\tag{5}$$
and compute the conserved current, factoring out $\omega$, from
$$j^{\mu}=\frac{\partial L}{\partial\left(\partial_{\mu}\phi\right)}\delta\phi-\delta L=-\phi\partial^{\mu}\phi\tag{6}$$
but something must be wrong because I can't show it is conserved,
$$\partial_{\mu}j^{\mu}=-\partial_{\mu}\left(\phi\partial^{\mu}\phi\right)=-\partial_{\mu}\phi\partial^{\mu}\phi-\phi\underset{=0\left(EL\right)}{\underbrace{\partial_{\mu}\partial^{\mu}\phi}=}-\partial_{\mu}\phi\partial^{\mu}\phi?\tag{7}$$
Can you please see what am I doing wrong?
 A: Why did you write $\frac{\partial L}{\partial (\partial_\mu\phi)}=-\phi\partial^\mu\phi$? 
It should be $=\partial^\mu\phi$.
update: everthing's correct, $\partial_\mu\phi\partial^\mu\phi$ vanishes because it is equal to $\partial_\mu\partial^\mu\phi$ up to a surface term. Whenever equation of motion is satisfied ($\partial_\mu\partial^\mu\phi=0$), $\partial_\mu\phi\partial^\mu\phi$ vanishes too.
A: 1) OP's action for a free scalar particle in $n$ spacetime dimensions is
$$ S~=~\int \!  \mathbb{L}(x), \qquad \mathbb{L}(x)~=~d^nx ~{\cal L}(x), \qquad  {\cal L}(x) ~:=~\frac{1}{2} \frac{\partial \phi(x)}{\partial x^{\mu}} \eta^{\mu\nu}\frac{\partial \phi(x)}{\partial x^{\nu}}.\tag{A}$$
It is not hard to check that the Lagrangian $n$-form $\mathbb{L}(x)$ is invariant under the scaling 
$$x^{\mu}\quad\longrightarrow\quad  x^{\prime\mu}~=~\lambda x^{\mu}, \qquad \phi(x)\quad\longrightarrow\quad  \phi^{\prime}(x^{\prime})~=~\lambda^{1-n/2} \phi(x) \tag{B}$$
with a positive parameter $\lambda > 0$. Of course, this is nothing but the well-known fact that in absolute units $\hbar=1=c$, the mass dimensions are
$$ [x]~=~ -1 , \qquad [S]~=~0, \qquad [\phi]~=~n/2-1. \tag{C}$$ 
It seems OP assumes that the spacetime dimension is $n=4$.
2) Next let us consider the corresponding infinitesimal transformation. Assume that $\lambda=1+\varepsilon$, where $\varepsilon$ is infinitesimal. The so-called horizontal infinitesimal variation is
$$\delta x^{\mu} ~:=~x^{\mu \prime}-x^{\mu} ~=~ \varepsilon \cdot x^{\mu}. \tag{D}$$ 
The infinitesimal variation of the dynamical variable $\phi$ is
$$ \delta \phi(x)~:= ~\phi^{\prime}(x^{\prime})-\phi(x)~=~\varepsilon \cdot (1-n/2) \phi(x), \tag{E}$$
so the vertical infinitesimal variation is 
$$ \delta_0 \phi(x)~:= ~\phi^{\prime}(x)-\phi(x)~=~\varepsilon \cdot (1-n/2 -x^{\nu}d_{\nu} )\phi(x).\tag{F}$$
In other words, the transformation (B) has horizontal generator $x^{\mu}$ and vertical generator $(1-n/2 -x^{\nu}d_{\nu} )\phi$.
3) The bare Noether current $j^{\mu}$ is defined as (the partial derivative of the Lagrangian density wrt. field derivatives) times the vertical generator plus the Lagrangian density times the horizontal generator:

$$ j^{\mu}~:=~ \frac{\partial {\cal L}}{\partial d_{\mu}\phi}\cdot(1-n/2 -x^{\nu}d_{\nu} )\phi+ {\cal L}\cdot x^{\mu}
~=~ d^{\mu}\phi\cdot(1-n/2 -x^{\nu}d_{\nu} )\phi
+ {\cal L}\cdot x^{\mu}.\tag{G} $$ 

It is straightforward to verify the on-shell continuum equation
$$ d_{\mu}j^{\mu}~\approx~0. \tag{H}$$
4) See also this related Phys.SE post. 
