Are Landweber exact spectra determined by their coefficient ring?

Question

Let $E$ be a Landweber exact ring spectrum. That is, we have a map of homotopy ring spectra $MU\rightarrow E$ and an isomorphism of homology theories $E_*X\simeq MU_*X\otimes_{MU_*}E_*$. Is the homotopy type of $E$ determined by the graded ring $E_*$?

My best guess is that the answer is "no" which would ideally be proved by finding a (graded) ring $R$ and two (graded), Landweber exact, formal group laws (graded ring maps) $e,f:MU_*\rightarrow R$, such that the spectra representing the two homology theories are not homotopy equivalent.

Answer 1

Put $R_*=\mathbb{Z}_{(p)}[x,y,y^{-1}]$ with $|x|=|y|=2$. Define $f,g\colon BP_*\to R_*$ by \begin{align*} f(v_i) &= \begin{cases} y^{p-1} & \text{ if } i = 1 \\ 0 & \text{ if } i > 1 \end{cases} \\ g(v_i) &= \begin{cases} x^{p-1} & \text{ if } i = 1 \\ y^{p^2-1} & \text{ if } i = 2 \\ 0 & \text{ if } i > 2 \end{cases} \\ \end{align*} This gives Landweber exact ring spectra $A_f$ and $A_g$ of heights one and two respectively, so for a finite spectrum $X$ of type $2$ we have $A_f\wedge X = 0 \neq A_g\wedge X$, so $A_f$ and $A_g$ are not equivalent as spectra.