I have the Klein-Gordon Lagrangian for three scalar fields and I want to find three independent infinitesimal transformations that leave the action invariant.
I suppose that these three transformations are rotation, translation and Lorentz transformation. But, if this first step is right I have no idea how to calculate the conserved currents because at least for translation and Lorentz transformation only the coordinates change and not the fields. Do you have any hint/idea?
This is fairly stardard QFT material: How to compute the Noether current for a general group of transformations that may involve both the fields and the coordinates $$ \left\lbrace \begin{aligned} x^\mu\to {x'}^\mu &= x^\mu + \varepsilon\,\xi^\mu(x)\,,\\ \phi^i(x) \to {\phi'}^i(x') &= \phi^i(x) + \varepsilon\,t^i[\phi](x)\,. \end{aligned} \right. $$ The operator $t^i[\phi]$ may be simply $t^i_{\,j}\phi^j$ but we can allow also for a more general functional dependence on $\phi$. The variation of the fields $\delta_\varepsilon\phi$ is (note the primes) $$ \delta_\varepsilon\phi(x) \equiv {\phi'}^i(x) - \phi^i(x) = \varepsilon \,\big(t^i[\phi](x) - \xi^\mu\partial_\mu\phi^i(x)\big)\,. $$ Asking for invariance amounts to the requirement that the Lagrangian transforms as a total derivative, so one must have $$ \delta_\varepsilon \mathcal{L} = \partial_\mu K_\varepsilon^\mu\,, \tag{1}\label{k} $$ for some $K_\varepsilon^\mu$. Define now the Noether current as $$ J_\varepsilon^\mu(x) = K_\varepsilon^\mu(x) - \frac{\delta\mathcal{L}}{\delta\partial_\mu\phi^i}\delta_\varepsilon\phi^i(x)\,.\tag{2}\label{j} $$ This is a Noether current because it is conserved when the equations of motion are satisfied. Indeed $$ \partial_\mu J^\mu_\varepsilon = \delta_\varepsilon\phi^i \frac{\delta\mathcal{L}}{\delta\phi^i} \approx 0\,. $$ The symbol $\approx$ means equality modulo stuff which vanishes when equations of motion are satisfied.
You can check for yourself that under Poincaré + dilatations the variations of the field are $$ \begin{aligned} &\mbox{Translations}:\\ &\;\;\;\delta_\epsilon \phi^i(x) = - \epsilon^\mu \partial_\mu \phi^i(x)\,,\qquad &&\epsilon^\mu\;\mbox{constant vector}\\ &\mbox{Lorentz}:\\ &\;\;\;\delta_\omega \phi^i(x) = - \tfrac12 \omega_{\mu\nu}\big((\Sigma^{\mu\nu})^i_{\phantom{i}j}\phi^j(x) + (x^{\mu}\partial^{\nu}-x^{\nu}\partial^{\mu})\phi^i(x)\big)\,,\qquad &&\omega_{\mu\nu}\;\mbox{antisymm. matrix}\\ &\mbox{Dilatations}:\\ &\;\;\;\delta_\lambda\phi^i(x) = - \lambda \big(x^\mu\partial_\mu +\Delta_i\big)\phi^i(x)\,,\qquad &&\lambda\;\mbox{real constant} \end{aligned} $$ The matrix $(\Sigma^{\mu\nu})^i_{\phantom{i}j}$ is the spin representation (e.g. $\frac14[\gamma^\mu,\gamma^\nu]$ for Dirac spinors, in your case it would just be zero because you have scalar fields) and $\Delta$ is the conformal dimension of $\phi$.
In order to compute the respective Noether currents you need to make the variation of the Lagrangian so as to compute $K^\mu$ as in \eqref{k} and then merely plug everything in \eqref{j}. The results will be of the form $$ J_{\epsilon}^\mu = \epsilon_{\nu} T^{\mu\nu}\,,\qquad J_{\omega}^\mu = \omega_{\nu\rho} \mathcal{S}^{\mu\nu\rho}\,,\qquad J_{\lambda}^\mu = \lambda \mathcal{D}^{\mu}\,. $$ So you can extract the current you need by basically dropping the $\epsilon_\mu,\,\omega_{\mu\nu}$ and $\lambda$. Just remember to antisymmetrize the last two indices of $\mathcal{S}^{\mu\nu\rho}$ before removing $\omega_{\nu\rho}$.
