Do an action and its Euler-Lagrange equations have the same symmetries? Assume a certain action $S$ with certain symmetries, from which according to the Lagrangian formalism, the equations of motion (EOM) of the system are the corresponding Euler-Lagrange equations.
Can it happen that the equations of motion derived by this procedure have different kinds and/or numbers of symmetries than the action one has started with? And if yes, are there underlying principles that state why which kind of symmetries the action does not have can emerge in corresponding EOMs or which kind of symmetries of the action can potentially disappear in the EOMs derived from the Euler-Lagrange equations?
 A: Setting. We are considering a transformation$^1$ that acts on the field variables $\phi^{\alpha}(x)$ and possibly the space-time point $x^{\mu}$. The transformation in turn apply to

*

*The action $S_V[\phi]=\int_V \! d^nx~{\cal L} $.


*The Euler-Lagrange equations = the equations of motion (EOM).


*A solution $\phi$ of EOM.
Definition. If any of the items 1-3 are invariant under the transformation, we speak of a symmetry of the corresponding item 1-3 .
Definition.  If a solution (3) doesn't have a symmetry that the EOM (2) have, we speak of a spontaneously broken symmetry.
Definition.  Next let us recall the definition of an (off-shell$^2$) quasi-symmetry of the action. It means that the action changes by a boundary integral
$$ S_{V^{\prime}}[\phi^{\prime}]
+\int_{\partial V^{\prime}} \!d^{n-1}x~(\ldots)
~=~S_V[\phi]+ \int_{\partial V} \!d^{n-1}x~(\ldots) \tag{0.1}$$
under the transformation.
Proposition.  In general, if an action (1) has a quasi-symmetry, then the EOM (2) must have a symmetry (wrt. the same transformation), cf. e.g. this Phys.SE post.
Examples:

*

*One example is the Maxwell Lagrangian density (in vacuum without the $J^{\mu}A_{\mu}$ source term)
$${\cal L} ~=~ -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}~=~\frac{1}{2}(\vec{E}^2-\vec{B}^2), \tag{1.1}$$
which doesn't have electromagnetic $SO(2,\mathbb{R})$ duality symmetry
$$(\vec{E}, \vec{B})\quad \longrightarrow \quad(\vec{E}\cos\theta - \vec{B}\sin\theta, \vec{B}\cos\theta + \vec{E}\sin\theta),\tag{1.2}$$
while the Euler-Lagrange equations (the Maxwell's equations in vacuum) are symmetric under electromagnetic duality.


*Another example is a non-relativistic free point particle where the Lagrangian
$$L~=~\frac{1}{2}m\dot{q}^2\tag{2.1}$$
is not invariant
under the Galilean symmetry
$$\dot{q}\quad \longrightarrow \quad\dot{q}+v,\tag{2.2}$$
nor the dilation/scale symmetry
$$ q \quad \longrightarrow \quad \lambda q,\tag{2.3}$$
but the EOM
$$\ddot{q}~=~0\tag{2.4}$$
is invariant. In the case of the Galilean symmetry (2.2), the Lagrangian changes by a total
time derivative
$$ L \quad \longrightarrow \quad L +mv\frac{d}{dt}\left( q +\frac{vt}{2}\right).\tag{2.5}$$
See also this Phys.SE post. Thus (2.2) is actually an example of a quasi-symmetry of the action. [It is an instructive exercise to derive the corresponding Noether charge $Q$. At the infinitesimal level, the Galilean transformation (2.2) reads
$$\begin{align} \delta \dot{q}~=~&\delta v~=~\varepsilon, \qquad \delta q~=~\varepsilon t,\cr \delta L ~=~& \varepsilon\frac{df}{dt}, \qquad f ~:=~mq. \end{align}\tag{2.6}$$
The bare Noether charge is
$$ Q^0~=~t \frac{\partial L}{\partial \dot{q}}~=~t m\dot{q},  \tag{2.7}$$
while the full Noether charge is
$$ Q~=~Q^0-f~=~m(\dot{q}t-q),\tag{2.8}$$
which is conserved on-shell, cf. Noether's Theorem. The (non-relativistic) Galilean boosts generator (2.8) should be compared to the (relativistic) Lorentz boosts generators $tP-xE$ in relativistic theories, cf. e.g. this Phys.SE post.]


*The dilation/scale transformation
$$ q \quad \longrightarrow \quad \lambda q,\tag{3.1} $$
is not a quasi-symmetry of the Lagrangian action
$$  S[q]~= ~\int\! dt ~L, \qquad L ~=~\frac{m}{2}\dot{q}^2- \frac{k}{2}q^2, \tag{3.2}$$
for the simple harmonic oscillator (SHO), but it is a symmetry of the EOM
$$ m\ddot{q}~=~-kq. \tag{3.3}$$


*The dilation/scale transformation
$$ q \quad \longrightarrow \quad \lambda q, \qquad  p \quad \longrightarrow \quad \lambda p, \tag{4.1}$$
is not a quasi-symmetry of the Hamiltonian action
$$\begin{align}  S_H[q,p]~= ~&\int\! dt ~L_H, \qquad L_H ~=~p\dot{q}-H, \cr H ~=~&\frac{p^2}{2m}+ \frac{k}{2}q^2,\end{align} \tag{4.2}$$
for the SHO, but it is a symmetry of Hamilton's EOM
$$ p~=~m\dot{q} , \qquad \dot{p}~=~-kq. \tag{4.3}$$


*The EOM of the SHO
$$  m\ddot{q}~=~-kq \tag{5.1}$$
is not invariant under the temporal symmetry
$$ t \quad \longrightarrow \quad \lambda t,\qquad \lambda~\neq~\pm 1,\tag{5.2}$$
but the trivial solution $q=0$ is.
--
$^1$ Note that in the main part of this answer the transformation acts only on the field variables $\phi^{\alpha}(x)$ and possibly the space-time point $x^{\mu}$, which is the type of transformation relevant for Noether's theorem. We are not considering a transformation of other objects (such as parameters) per se.
Example of the latter: A transformation of the Lagrangian density $${\cal L} \longrightarrow \lambda {\cal L},\qquad  \lambda~\neq~ 1,\tag{6.1}$$
is not a quasi-symmetry of the Lagrangian density, but it is a symmetry of the EL equations.
$^2$ Here the word off-shell indicates that the EOM are not assumed to hold under the specific transformation. In case of continuous transformations, if we assume the EOM to hold, then any infinitesimal variation of the action is trivially a boundary integral.
