= Solution
Take a constant infinitesimal <translation> with fixed-coordinate <field variation> $\delta\phi=\epsilon^\nu\partial_\nu\phi$. The associated change in the <Lagrangian density> is $\delta\mathcal L=\epsilon^\nu\partial_\nu\mathcal L=\partial_\mu(\epsilon^\nu\delta^\mu{}_{\nu}\mathcal L)$. Substituting this total derivative into the <Noether current> gives $J^\mu=\epsilon^\nu T^\mu{}_{\nu}$, where
$$
\boxed{T^\mu{}_{\nu}=\partial^\mu\phi\,\partial_\nu\phi-\delta^\mu{}_{\nu}\left[\frac12\partial_\rho\phi\,\partial^\rho\phi-\frac12m^2\phi^2-\frac{\lambda}{24}\phi^4\right].}
$$
This is the <canonical stress-energy tensor>. The chosen sign of the <translation> parameter produces precisely this sign of the <Noether current>; reversing the parameter reverses the current without changing its conservation law.
To check <current conservation> directly, differentiate the <tensor>. The terms containing $\partial_\nu\partial_\rho\phi$ cancel against the derivative of the <kinetic term>, leaving
$$
\partial_\mu T^\mu{}_{\nu}=
\left(\Box\phi+m^2\phi+\frac{\lambda}{6}\phi^3\right)\partial_\nu\phi
=F\partial_\nu\phi.
$$
Thus $\partial_\mu T^\mu{}_{\nu}=0$ <on shell>. Raising the second <Lorentz index> gives $T^{\mu\nu}=T^{\nu\mu}$: this <scalar field> already has a symmetric <canonical stress-energy tensor>.
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