Scalar Field Redefinition and Scattering Amplitude Consider a field redefinition
$$
\phi \rightarrow \phi' = \phi+\lambda \phi^2
$$
Find the Feynman rules for this theory and work out the $2\rightarrow 2$ scattering amplitude at tree level (The result should be zero).
$$
\mathcal{L}_0 = -\frac{1}{2}(\partial^\mu \phi \partial_\mu \phi + m^2 \phi^2) \implies \text{EOM: } \Box\phi - m^2\phi=0 
$$
Preforming the field redefinition on the Lagrangian I obtain:
$$
\mathcal{L}_0 \rightarrow \mathcal{L}'= \mathcal{L}_0 -2\lambda\phi \partial^\mu \phi \partial_\mu \phi - 2 \lambda^2 \phi^2 \partial^\mu \phi \partial_\mu \phi -\lambda m^2 \phi^3 - \frac{\lambda^2}{2} m^2 \phi^4
$$
I then rewrote the two terms with partial derivatives as
$$
-2\lambda\phi \partial^\mu \phi \partial_\mu \phi = -\lambda \partial^\mu (\phi^2 \partial_\mu \phi)+\lambda \phi^2 \Box \phi
$$
which gets rid of the total derivative.  Then I used the EOM to replace $\Box \phi$ with $m^2 \phi$ to get
$$
\mathcal{L}' = \mathcal{L}_0 + \lambda (m^2 - 1)\phi^3 + \lambda^2 \left( \frac{m^2}{3}-\frac{1}{2} \right) \phi^4
$$
Now this looks like a normal scalar field theory with a cubic and quartic vertex.  Now it wants  a tree level diagram for two to two scattering, so I should put these two vertices in all the combinations that have no loops and result in two in, and two out?  Then subsequently work out the scattering amplitude from the said diagrams using the Feynman rules I put together with the above Lagrangian?
 A: I remember this as a nice problem from chapter 10 of Srenicki's QFT book. Your first expression for $\mathcal{L}'$ should have factors of $m^2$ in the last two terms. You should work with the lagrangian from here and derive the feynman rules and calculate the scattering amplitude. I will sketch this for you below. It is a very nice excercise to check whether you have understood everything up to that point.
You cannot use the EOM as you have done since that was for the original $\phi$ before the field redefinition. After the field redefinition your new $\phi$ (may help you to call it something else, e.g. $\phi \to \psi + \lambda \psi^2$) has a different equation of motion involving quadratic and cubic terms!
As I said above you should just work from your first expression for  $\mathcal{L}'$ with the derivatives in and derive the Feynman rules from there. If you go through the whole procedure as is done in the previous chapter (being careful about the derivatives and symmetry factors) for each type of vertex (from each new term in the lagrangian) you will find a whole bunch of new vertex factors. Just calculate the 3 or 4 point function for each individual term and determine the feynman rules that way. Once you get used to how it works you can start to read them off directly from the lagrangian but it is good to go through the whole procedure a few times first. You should find for example, the cubic term with no derivatives gives a vertex joining three lines with a vertex factor of
$$-i3! \lambda m^2.$$
The cubic vertex with derivatives has a vertex factor of
$$4i\lambda(k_1k_2 + k_1k_3 + k_2k_3)$$
you see that the derivatives act to pull down powers of the momenta from the Fourier transforms of the propagators. It is sort of obvious if you just think about $\partial_\mu$ in momentum space.
So the total vertex joining three lines will just be the sum of these two. Do the same for the four point vertex and then as you say, draw all the feynman diagrams for $\phi\phi \to \phi\phi$ scattering, calculate the total amplitude and you should find if you do everything right that you get lots of nice cancellations and the answer is zero and you will have a nice warm feeling inside :-)
Derivative interactions such as these crop up in non-abelian gauge theories (e.g. the standard model) and gravity so they are important to understand. If you are still having trouble calculating the vertex factors I highly recommend this pdf which goes through in painstaking detail how to derive feynman rules. It follows the treatment by Srednicki too! Check section 5.3 and 5.4 for what you need.
A: I guess that there are mistakes in your algebra. In any case:

Now this looks like a normal scalar field theory with a cubic and
  quartic vertex

Except for the fact that there are just two independent parameters $m$ and $\lambda$.
The cross section must be zero just because of the LSZ reduction formula and the last Lagrangian should be related to the first, which has no interactions, by a field redefinition. In deduction of the LSZ formula, it is clear that two Lagrangians related by a field redefinition give the same S-matrix elements regardless the off-shell n-point function are different in each case. The only condition is that both fields have the same matrix elements between the vacuum state and one particles states and the same vacuum expectation value. Matrix elements between the vacuum state and multiparticle states do not matter because of the Riemann-Lebesgue lemma and how these matrix elements enter in the LSZ formula.   
A: See 't Hooft and Veltman's diagrammar lecture notes. There's a discussion of something called the 'equivalence theorem'. Essentially, it observes that since the S-matrix is obtained from the correlation function by finding the residue at a single-particle-pole, everything which would lead to multiple particle poles does not contribute. There's a loophole if the field transformation is not local, but leads to a one-particle pole.
