Firstly, one should distinguish the Lagrange function from the Lagrangian density and the action.
The Lagrange density
The Lagrange density $\mathcal L$ is a function with imput values $(\phi, \partial \phi, x) \mapsto \mathcal L (\phi, \partial \phi, x)$ (Technically speaking, the Lagrange density is a function on some jet space, but this does not really matter here.) the Lagrange function $L$ and the the action $\mathcal S$.
Let us first regard $\phi$ and $\partial \phi$ not as fields and derivatives thereof but as abstract coordinates (of jet space). This is similar to $q$ and $p$ in phase space. A Lagrange density can be translation invariant in the sense, that does not explicitly depend on $x$ i.e.,
$$ \mathcal L (\phi, \partial \phi, x ) = \mathcal L (\phi, \partial \phi, x' )\tag{1}$$ for all $x'$.
Given a representation of the Lorentz group $\rho: SO(1,3)\times V \to V $ that acts on $\phi \in V$ we can ask for Lorentz invariance of $\mathcal L$ in the following sence $$\mathcal L(\rho(\Lambda) \phi, \tilde\rho(\Lambda)\partial \phi, x) = \mathcal L(\phi, \partial \phi, x)\,\tag{2}.$$ ( Here $\tilde \rho$ is an induced representation. E.g., if $\partial \phi = (\phi_{,\mu}, \phi_{,\alpha \beta}, \dots)$ the entries transform like partial derivatives of a $V$ valued function. E.g, $\phi_{,\mu} \mapsto \Lambda^{-1}{}^{\nu}{}_{\mu}\rho(\Lambda) \phi_{, \nu}$.)
These two properties can usually be checked easily by some index calculus.
Now let's look at a fields $\phi: \mathbb R^{1+3} \to V, x \mapsto \phi(x)$. We can define the following map
$$x \mapsto \mathcal L ( \phi(x), \partial \phi(x), x)$$
As the field $\phi$ is usually not Lorentz invariant ( $\phi(x) \neq \rho(\Lambda) \phi( \Lambda^{-1} x)$), above map is also not Lorentz invariant. I.e., $$ \mathcal L(\phi(\Lambda^{-1} x), \partial \phi(\Lambda^{-1} x), \Lambda^{-1} x) = L(\rho(\Lambda)\phi(\Lambda^{-1} x), \tilde \rho(\Lambda)\partial \phi(\Lambda^{-1} x), x) \neq \mathcal L(\phi(x), \partial \phi(x), x)\,.$$
The Lagrange function
Given some preferred time coordinate $t = x^0$ you can defined the Lagrange function by integrating over the spacelike dimensions $$ L(t) := \int_{\mathbb R^3} d^3 x \mathcal L(\phi(\vec x , t),\partial \phi (\vec x, t), \vec x, t)\,.$$
It can also be seen as functional $L(\phi, \partial \phi,t)$, where $\vec x \to \phi(\vec x)$ and $\vec x \to \partial(\vec x)$ are functions defined on space (technically they might need to satisfy certain constraint equations, e.g., the space like partial derivatives need to satisfy $\partial_a \phi(\vec x) = \phi_{,a}(x)$).
This functional is also not Lorentz invariant. It makes no sense to Lorentz transform $\vec x$ and therefore it makes no sense to Lorentz transform a field defined on space (and not space time) such as $\vec x \mapsto \phi(\vec x)$.
However, if $\mathcal L$ satisfies $(1)$ then $L(\phi, \partial \phi, t)$ does not explicitly depend on time and it is space translation invariant. I.e., if $\phi'(\vec x ) := \phi(x- \Delta x)$, $\partial \phi'(\vec x):= \partial \phi (x- \Delta x)$ then $$ L(\phi', \partial \phi', t) = L(\phi, \partial \phi, t)$$
Further, if $\mathcal L$ satisfies $(2)$, $L$ is invariant under translation in space and rotation in space, i.e., the Euclidean subgroup of the Poincaré transformations. That is, for $ \phi'(\vec x) := \rho(R)\phi(R^{-1}\vec x)$ and $\partial \phi'(\vec x) = \tilde \rho(R))\partial \phi(R^{-1} \vec x)$, we have $$ L(\phi', \partial \phi', t) = L(\phi, \partial \phi, t)$$
Both of these cases can be seen by using appliyng the transformations formula to the defining integral.
The action
the action is defined as
$$\mathcal S(\phi) := \int_{\mathbb R^{1+3}} dx^4 \mathcal L(\phi(x), \partial \phi(x), x) = \int_{\mathbb R} dt L(\phi(\cdot, t),\partial \phi(\cdot, t), t)\,.$$
If $\mathcal L$ satisfies $(1)$ and $(2)$ it follows from the integration formula that form $\phi'(x) = \rho(\Lambda )\phi(\Lambda^{-1}x - \Delta x)$ we have $$\mathcal S(\phi) = \mathcal S(\phi')$$. This is a consequence of the transformation formula for integrals.
