In a (classical) lagrangian field theory, the configuration space $\mathcal C$ of the system is a space of field configurations. A field configuration (or just "field" for short) is usually taken to be a function $\phi:\mathcal M\to T$ where $M$ is a manifold and $T$ is some set, often a manifold or vector space or both, called the target space of the field. The configuration space $\mathcal C$ is then taken to be some sufficiently smooth (when a notion of smoothness can be defined) subset of the set of all possible fields. The lagrangian is then a function $L:\mathcal C\to\mathcal C$;
\begin{align}
\phi\mapsto L[\phi]
\end{align}
Namely, the lagrangian maps a particular field configuration, to another field configuration. Often, one considers a field theory for which the lagrangian can be written as local local density, but this is not strictly speaking necessary. The action of the theory can then be defined as the integral of $L[\phi]$ over $M$;
\begin{align}
S[\phi] = \int_M \, d^Dx\,L[\phi](x).
\end{align}
Note. My terminology and notation here are a bit non-standard in some contexts. For example, in relativistic physics (field theory) the Lagrangian will usually map a field configuration $\phi$ to a function $L[\phi]$ of time, and then this function of time will be integrated to yield the action. It's not hard in practice to translate between conventions.
One can then define what it means for the action to possess symmetry. In particular, given a mapping $F:\mathcal C\to \mathcal C$ of the manifold on which field configurations are defined to itself, one says that the action is invariant under $F$ provided
\begin{align}
S[F(\phi)] = S[\phi]
\end{align}
for all $\phi\in\mathcal C$. For "continuous" transformations, one can also define notions of symmetry that don't involve full invariance, but let's keep the discussion simple at this point.
Example. A common toy theory considered as the first example in most relativistic field theory texts is that of a single, free, real Lorentz scalar defined on Minkowski space (I'll use metric signature $+ - - -$). In this case, we have
\begin{align}
M = \mathbb R^{3,1}, \qquad T = \mathbb R
\end{align}
and $\mathcal C$ is a space of sufficiently smooth functions $\phi:\mathbb R^{3,1}\to\mathbb R$ that satisfy certain desired boundary conditions. The Lagrangian of such a theory is
\begin{align}
L[\phi](x) = \mathscr L(\phi(x), \partial_0\phi(x), \dots \partial_3\phi(x))
\end{align}
where $\mathscr L$ is the Lagrangian density defined by
\begin{align}
\mathscr L(\phi, \partial_0\phi, \dots, \partial_3\phi)
&= \frac{1}{2}\partial_\mu\phi\partial^\mu\phi - \frac{1}{2}m^2\phi^2.
\end{align}
Given any Lorentz transformation $\Lambda$, one can define a transformation $F_\Lambda:\mathcal C\to\mathcal C$ as follows:
\begin{align}
F_\Lambda(\phi)(x) = \phi(\Lambda^{-1} x).
\end{align}
A short computation then shows that the Lagrangian is a Lorentz scalar under this transformation, namely
\begin{align}
L[F_\Lambda(\phi)](x) = L[\phi](\Lambda^{-1}x).
\end{align}
In fact, this is essentially done for you on page 36 of Peskin. It follows from this that the action is invariant under $F$;
\begin{align}
S[F_\Lambda(\phi)] = \int_{\mathbb R^{3,1}} d^4x\, L[\phi](\Lambda^{-1}x) = \int_{\mathbb R^{3,1}} d^4x\, L[\phi](x) = S[\phi].
\end{align}
since the measure $d^4 x$ is Lorentz-invariant. Notice, in particular, that the fact that the Lagrangian transformed as a Lorentz scalar (namely precise in the same way as the scalar field $\phi$ was defined to transform), immediately led to invariance of the action.
Furthermore, suppose that $\phi$ is a field configuration that leads to stationary action, then we can also show that a Lorentz-transformed field leads to a stationary action using Lorentz invariance of the action. To see this, recall that the variational derivative in the direction of a field configuration $\eta$ is defined as follows:
\begin{align}
\delta_\eta S[\phi] = \frac{d}{d\epsilon}S[\phi+\epsilon\eta]\Big|_{\epsilon=0}
\end{align}
Now, suppose that $\phi$ is a stationary point of the action, namely that $\delta_\eta S[\phi] = 0$ for all admissible $\eta$, then for all such $\eta$ we have
\begin{align}
\delta_{F_\Lambda(\eta)} S[F_\Lambda(\phi)]
&= \frac{d}{d\epsilon} S[F_\Lambda(\phi) + \epsilon F_\Lambda(\eta)]\Big|_{\epsilon = 0} \\
&= \frac{d}{d\epsilon} S[F_\Lambda(\phi+\epsilon_\eta)]\Big|_{\epsilon = 0} \\
&= \frac{d}{d\epsilon} S[\phi+\epsilon_\eta]\Big|_{\epsilon = 0} \\
&= 0
\end{align}
Now set $\eta = F_\Lambda^{-1}(\xi)$, then the computation we just performed shows that
\begin{align}
\delta_{\xi} S[F_\Lambda(\phi)] =0.
\end{align}
for all admissible field configurations $\xi$. In other words, the Lorentz transformed scalar is also a stationary point of the action. Notice that this demonstration holds for any Lorentz transformation, not just boosts.
Addendum.
As pointed out in the comments, the argument at the end about variational derivatives hinges on linearity of $F_\Lambda$. This can be demonstrated as follows:
\begin{align}
F_\Lambda(a\phi+b\psi)(x)
&= (a\phi+b\psi)(\Lambda^{-1}x) \\
&= a\phi(\Lambda^{-1}x) + b\psi(\Lambda^{-1}x) \\
&= aF_\Lambda(\phi)(x) + bF_\Lambda(\psi)(x).
\end{align}
Let me make some remarks about the mapping $F:\mathcal C\to \mathcal C$; a symmetry of the action. If there exists a mapping $f_T:T\to T$ on the target space that induces this mapping, namely
\begin{align}
F(\phi)(x) = f_T(\phi(x)),
\end{align}
then $F$ is called an internal symmetry. On the other hand, if there is a mapping $f_M:M\to M$ on the base manifold $M$ that induces this mapping, namely
\begin{align}
F(\phi)(x) = \phi(f_M(x)),
\end{align}
then $F$ is called a base manifold symmetry (or more commonly a spacetime symmetry since in the context of relativistic field theory, the base manifold is a spacetime like Minkowski space.)
Furthermore, the mapping $F:\mathcal C \to\mathcal C$ on the field configuration space is often, as in the scalar field example, a group action of some group $G$ on $\mathcal C$. This means that to each $g\in G$, we associate a mapping $F_g:\mathcal C\to \mathcal C$ such that the mapping $g\mapsto F_g$ is a homomorphism of the group $G$. In practice, the group $G$ is sometimes a group of symmetries that naturally acts on the base manifold, and sometimes $G$ a group of symmetries that naturally acts on the target space (or even both when the $M=T$). In any event, this group action is usually obtained by composing a target space group action $(f_T)_g:T\to T$ with a base manifold group action $(f_M)_g:M\to M$. More explicitly, for each $g\in G$, we can define mappings $(F_T)_g:\mathcal C\to \mathcal C$ and $(F_M)_g:\mathcal C\to\mathcal C$ as follows:
\begin{align}
(F_T)_g(\phi)(x) = (f_T)_g(\phi(x)), \qquad (F_M)_g(\phi)(x) = \phi((f_M)_g(x))
\end{align}
and then the full group action $F_g:\mathcal C\to\mathcal C$ is defined by the composition of these two;
\begin{align}
F_g = (F_T)_g\circ (F_M)_g
\end{align}
or more explicitly
\begin{align}
F_g(\phi)(x) = (f_T)_g(\phi((f_M)_g(x))).
\end{align}
Now, this is all a bit abstract, so let's write out what all of these objects would be for the scalar field example:
\begin{align}
G &= \mathrm{SO}(3,1) \\
g &= \Lambda\\
(f_T)_g(\phi(x)) &= \phi(x) \\
(f_M)_g(x) &= \Lambda^{-1}x \\
(F_T)_g(\phi)(x) &= \phi(x) \\
(F_M)_g(\phi)(x) &= \phi(\Lambda^{-1}x) \\
F_g(\phi)(x) &= \phi(\Lambda^{-1}x)
\end{align}
Notice that $f_T$ is simply the identity mapping on the target space. This is precisely what we mean when we say that the scalar field is a Lorentz scalar. On the other hand, for a Lorentz vector, the target space itself would be Minkowski space $\mathbb R^{3,1}$, and the target space group action would be
\begin{align}
(f_T)_\Lambda(A(x)) = \Lambda A(x),
\end{align}
namely, there is an internal symmetry in which the vector indices on the field transform non-trivially. In components, which is how you'll see this written in Peskin for example, the right hand side would be written as $\Lambda^\mu_{\phantom\mu\nu} A^\mu(x)$.
In case of non-field quantity that has one value for the whole inertial system, like net electric charge of a body, it means its value is the same in all inertial systems. For example, electron has the same charge in all inertial systems. Therefore it is Lorentz invariant.
In case of a field quantity like $E^2 - c^2B^2$, the value depends on position and time (event). In one inertial system the value of this quantity for event $x^\mu$ is
$$
E^2(x^\mu) - c^2B^2(x^\mu)
$$
In another system where the same event has coordinates $x'^{\mu}$ and the electric and magnetic fields are given by functions $\mathbf E',\mathbf B'$, the value is
$$
E'^2(x'^\mu) - c^2B'^2(x'^\mu).
$$
It can be shown that
$$
E^2(x^\mu) - c^2B^2(x^\mu) = E'^2(x'^\mu) - c^2B'^2(x'^\mu).
$$
It is this property that is meant when saying $E^2-c^2B^2$ is Lorentz invariant. In general case, a field $\phi(x^\mu)$ is Lorentz invariant if its evaluation in two inertial systems, connected via Lorentz transformation, leads to the same value:
$$
\phi(x^\mu) = \phi'(x'^\mu).
$$
Best Answer
Classical field theory
Let's start with classical field theory of a scalar field, $\phi(x^\mu)$, where $x^\mu$ are coordinates on spacetime. Then all dynamics are derivable from the action via the Euler-Lagrange equations. The action typically takes the form \begin{equation} S = \int {\rm d}^4 x \mathcal{L}(\phi,\partial\phi) \end{equation} Under a Lorentz transformation $\Lambda$, $\phi(x)\rightarrow\phi(\Lambda^{-1}x)$. If the action remains invariant when we perform this transformation, then the equations of motion will be invariant under this transformation. As a result, if $\phi(x)$ obeys the equations of motion, then so will $\phi(\Lambda^{-1}x)$.
Crucially, we can apply a Lorentz transformation to $\phi$, whether or not the action is invariant under a Lorentz transformation. The notion of invariance, or symmetry, has two parts: (1) we apply the transformation, and (2) the action does not change when we perform the transformation.
An example of an action which will be invariant under Lorentz transformations is
\begin{equation} S = \int {\rm d^4 x} \left(-\frac{1}{2} \eta^{\mu\nu}\partial_\mu \phi \partial_\nu \phi - V(\phi)\right) \end{equation} where I've chosen the metric signature $\{-1,+1,+1,+1\}$. Meanwhile, an action which is not Lorentz invariant is \begin{equation} S = \int {\rm d^4 x} \left(-\frac{1}{2} \eta^{\mu\nu}\partial_\mu \phi \partial_\nu \phi + V(x)^\mu \partial_\mu \phi)\right) \end{equation} where $V(x)^\mu$ is some 4-vector that depends on space. For example, we could take $V(x)^\mu=x^\mu$. To check the action is not Lorentz invarant, we can pefrom the transformation $\phi(x)\rightarrow\phi(\Lambda^{-1}x)$ and simply check that the action changes. This shows that the transformation law, by itself, is not enough to guarantee the theory is invarant.
Quantum field theory (canonical formulation)
The story is similar, but richer, at the quantum level. If we take a path integral approach, then the above story is (almost) sufficient quantum mechanically (with the subtlety being that we have to check if the path integral measure is also invariant, after renormalization).
In the canonical formalism, $\phi(x)$ is an operator valued distribution on spacetime. The equation $U(\Lambda)^\dagger \phi(x) U(\lambda) = \phi(\Lambda^{-1}x)$ explains how to represent Lorentz transformations of the field on Hilbert space.
In "vanilla" non-relativisitc quantum mechanics, to check if a given transformation $T$ is a symmetry, we represent the transformation $T$ via unitary operators $U(T)$ and see if the Hamiltonian is invariant, $U^\dagger(S)HU(S)=H$.
In our relativistic case, we have to be more careful, since boosts will change the Hamilonian. What we really need is that the Lorentz transformations act on the energy-momentum 4-vector as
\begin{equation} U^\dagger(\Lambda) P^\mu U(\Lambda) = \Lambda^\mu_{\ \ \nu}P^\nu \end{equation}
where $P^\mu=\{H,P^i\}$, and $H$ is the Hamiltonian and $P^i$ are the spatial components of the momentum. We also require that the angular momentum tensor $M^{\mu\nu}$ transforms as a tensor \begin{equation} U^\dagger(\Lambda) M^{\mu\nu} U(\Lambda) = \Lambda^\mu_{\ \ \alpha} \Lambda^\nu_{\ \ \beta} M^{\alpha\beta} \end{equation} The quantities $P^\mu$ and $M^{\mu\nu}$ can be derived by using Noether's theorem, applied the action, for spacetime transformations and Lorentz transformations, respectively. Explicit forms for the scalar field can be found, for example, in Chapter 1 of David Tong's QFT lecture notes, http://www.damtp.cam.ac.uk/user/tong/qft.html.
In practice, one often works with an infintesimal version of these laws. For a translation, we write $U(\Lambda)=1+i a^\mu P_\mu$, where $a^\mu$ is the 4 vector defining the translation, and $P_\mu$ is the "generator" of the transformation, which we identify with the Hamiltonian and momentum as above. For a Lorentz transformation (boosts and rotations), $U(\Lambda)=1+i \omega^{\mu\nu} M_{\mu\nu}$, where $\omega^{\mu\nu}=-\omega^{\nu\mu}$ are the parameters of the Lorentz transformation, and $M_{\mu\nu}$ are the generators. Then the group transformation laws above, imply the following commutation rules for the generators
\begin{eqnarray} [P_\mu,P_\nu]&=&0 \\ [M_{\mu\nu},P_\rho]&=& -i \left( \eta_{\mu\rho} P_\nu - \eta_{\nu \rho} P_\mu\right)\\ [M_{\mu\nu},M_{\rho\sigma}]&=&-i\left(\eta_{\mu \rho}M_{\nu\sigma} - \eta_{\mu \sigma}M_{\nu\rho}-\eta_{\nu\rho}M_{\mu\sigma}+\eta_{\nu\sigma}M_{\mu\rho}\right) \\ \end{eqnarray}
So, the steps for checking whether a given field theory has a Lorentz symmetry in canonical quantization are:
A few additional notes:
Often, if the action is Lorentz invariant, then the quantum theory will be as well.
Sometimes this logic is inverted and we start from the commutation relations and construct a theory which obeys the necessary relationships.
In classical field theory, we can formulate the question of symmetry in a very similar way using Poisson brackets.
The approach in this answer was to find a generalization of the idea from non-relativistic quantum mechanics that "a symmetry leaves the Hamiltonian invariant." However, we can take a more abstract point of view as well, in that we want to have a unitary representation of the Poincair{'e} group on Hilbert space. The Noether charges $P_\mu$ and $M_{\mu\nu}$ provide the generators for this representation.
To come back to your question: the Lorentz transformation law, by itself, does not imply the theory has Lorentz symmetry. However, it is an important ingredient in checking if a given theory is Lorentz invariant. And, the fact that we were able to construct a unitary representation of the Poincair{'e} group on Hilbert space, is equivalent to having the theory being symmetric.