Symmetry of Euler-Lagrange equations and conservation laws 
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*Continuous symmetry of the action implies a conservation law, but what if equations of motion have a continuous symmetry? Does it imply a conservation law? 

*Also is symmetry of equations of motion also a symmetry of the action? 
I assume that the equations of motion I am talking about follow from the action, i.e. the equations of motion are Euler-Lagrange equations. Related examples could help and references also.
 A: Different Lagrangian could give the same equation of motion. If you add non-dynamical variables, you could present a Lagrangian without symmetries, while the equation of motion has symmetries.
For instance, take the Lagrangian : 
$$L(x,f) = \frac{f^2}{2}+\dot f ~x$$
Euler-Lagrange equations (applied to $x$ and $f$) give : 
$$\dot f = 0, f = \dot x$$
That is : 
$$\ddot x=0,  f = \dot x$$
The Lagrangian does not respect explicitely translational $x$-invariance (because of the term $x$), but equations of movement respect translational $x$-invariance, and $\dot x$ is a conserved quantity.
$f$ is non-dynamical, because we can replace $f$ by its value due to the equation of movement and get a Lagrangian :
$$L(x) = \frac{\dot x^2}{2}$$
With applications of Euler-Lagrange equation, we obviously get the same equation of movement $\ddot x = 0$. This latter lagrangian obviously respects the translational $x$-invariance.
A: You cannot prove the existence of a conserved current directly from the equations of motion in general just given the fact that the equations of motion have a continuous symmetry.
A counterexample to this is a damped harmonic oscillator, $\ddot{x}+\Gamma \dot{x}+\omega^2 x=0.$ This is invariant under $t\rightarrow t+\delta t$, but doesn't have a conserved quantity.
Now, given a specific equation of motion, you can look for constants of motion. For example, given $\ddot{x}=V'(x)$ you can prove that energy is conserved. For example see First integral of an equation of motion: $\mu\ddot r=-\frac{k}{r^2}$. But it's hard to see how this method relates to any particular symmetry of the eom, rather than the particular algebraic structure of the eoms.
To relate symmetries to conserved quantities, you should use the action formulation, as you know. Symmetries of the action are always symmetries of the equations of motion. However, not all equations of motion have a corresponding action (like the damped harmonic oscillator). There is no noether's theorem that operates directly at the level of the eoms, that's why I keep going back to the action.
It seems you want to know about the other way around, are symmetries of the eoms always symmetries of the action? I don't see how this could fail to be the case, but admittedly I can't prove it off the top of my head. Here's why I think that symmetries of the eoms should be symmetries of the action: Let's say the symmetry is translation invariance just for simplicity, you could generalize this argument to any symmetry easily. Let's say the eoms are translation invariant. Then I can set up my system at position $A$, let it evolve classically, and see what happens. Then I move my system to be position $B$ and set it up with identical initial conditions. Exactly the same motion must happen at position $B$ by translation invariance. But the action is just a function of the path taken--you would need to get a different number for the action evaluated on the paths taken at position $A$ then you would get if you evaluated the action on the paths at position $B$. But since the physics doesn't distinguish between $A$ and $B$ how could this be true: by simply calculating the action you could tell whether you were at $A$ or $B$. Admittedly it isn't a proof, but it's why I don't think it's true. Even if you could find a symmetry of the eoms that wasn't a symmetry of the action, you wouldn't be able to use Noether's theorem to show there was a conserved quantity, so there wouldn't be a conserved quantity associated with that symmetry.
Not all conservation laws follow from symmetries though.
