Let $\mathfrak A$ and $\mathfrak B$ be finite linears orders and $m<\omega$. Then we have $$\mathfrak A \equiv^m_{FO}\mathfrak B \quad\text{iff}\quad |A|=|B| \,\, \text{or}\,\, |A|,|B|\ge 2^m-1.$$
My question: Is there a similar result for monadic second order logic?
So I'm looking for a result like: $$\mathfrak A \equiv^m_{MSO}\mathfrak B \quad\text{iff}\quad ?$$
My thoughts concerning this question:
There are only finitely many theories of bounded quantifier rank in MSO. So although MSO has a stronger expressive power than FO there are different finite linear orders which are not distinguishable via MSO given a certain quantifier rank. Nevertheless, for large quantifier ranks $m<\omega$ there won't be a $n<\omega$, such that all orders with at least $n$ elements have the same theory. This is because there is a MSO formular $\varphi$ which is true iff the order has even cardinality.
$$\varphi:= \exists X \forall x [(\min(x)\rightarrow Xx) \land (\max(x)\rightarrow \lnot Xx) \land \forall y (\text{succ}(x,y)\rightarrow (Xx\leftrightarrow \lnot Xy)) ]$$ with
$$\min(x):= \forall y \, x\le y$$ $$\max(x):= \forall y \, y\le x$$ $$\text{succ}(x,y):= \forall z [(x\le z \land z\le y)\rightarrow (x=z \lor y=z)]$$
This $\varphi$ has quantifier rank 4, so for $m\ge 4$ there won't be a $n<\omega$, such that all finite orders $\mathfrak A$ with at least $n$ elements have the same theory $\text{Th}_{MSO}^m(\mathfrak A)$. In particular we see that all orders of the same theroy are of the same parity.