Main question
Consider the stochastic process $\{x_t\}_{t\in T}$ with $T=\mathbb{Z}_+$ on $(\Omega,\mathcal{F},\mathbb{P})$.
If we let $x$ be adapted to the natural filtration $\{\mathcal{F}_t\}_{t\in T}$ we can ask questions about the law, $\mathbb{P}^x$, like whether it is a Martingale w.r.t. it etc. and so write down expressions like
$$\mathbb{P}^x_{n+1}(x_{n+1}|\mathcal{F}_n)=\mathbb{P}^x_{n+1}(x_{n+1}|\sigma(x_m:m\leq n))$$
where, as usual, $\mathcal{F}_n\subseteq\mathcal{F}_{n+1}$ and, given some simple case where the measureable space for $x$ is finite such that it takes values in $A$ with sigma algebra $2^A$ we can write down finite dimensional distributions of $\mathbb{P}^x$ which are going to look something like a chapman kolmogorov eq
$$\mathbb{P}^x_{0,\ldots,n}(x_0,x_1,x_2,\ldots,x_n)=\mathbb{P}^x_0(x_0)\mathbb{P}^x_1(x_1|\mathcal{F}_0)\ldots\mathbb{P}^x_n(x_n|\mathcal{F}_{n-1})\\ =\mathbb{P}^x_0(x_0)\mathbb{P}^x_1(x_1|\sigma(x_0))\ldots\mathbb{P}^x_n(x_n|\sigma(x_0,x_1,\dots,x_{n-1}))$$
To be Markov we have $\mathbb{P}^x_{n+1}(x_{n+1}|\mathcal{F}_n)=\mathbb{P}^x_{n+1}(x_{n+1}|\sigma(x_n))$ but I want to think about this the other way around.
If under the natural filtration $x$ is non-markov, I want to consider a law on $x$ where it is Markov, i.e. that it uses only information from the previous time step. This means that I am asserting the existence of some measure $\mathbb{Q}^x$ such that
$$\mathbb{Q}^x_{0,\ldots,n}(x_0,x_1,x_2,\ldots,x_n)=\mathbb{Q}^x_0(x_0)\mathbb{Q}^x_1(x_1|\sigma(x_0))\mathbb{Q}^x_2(x_2|\sigma(x_1))\ldots\mathbb{Q}^x_n(x_n|\sigma(x_{n-1}))$$
The question is can I consider this law $\mathbb{Q}^x$ as arising from the process $x$ being adapted to some 'non-natural' filtration $\mathcal{F}_t^M$ where $\mathcal{F}_n^M=\sigma(x_{n-1})$? The first and obvious issue is that we pretty emphatically don't have $\mathcal{F}_n^{M}\subseteq \mathcal{F}_{n+1}^{M}$. Can this be cast in such a way? Is this even a sensible question?
Quick secondary question
Say we have two stochastic processes, $\{x^1_t\}_{t\in T}$ $\{x^2_t\}_{t\in T}$, am I correct in saying that the natural filtration would be $\mathcal{F}_n=\sigma(x^1_m,x^2_m:m\leq n)$ and that this would be different to the natural filtration on just $x^1$ which would be $\mathcal{F}^1_n=\sigma(x^1_m:m\leq n)$? And that we would have $\mathcal{F}^1_n\subseteq \mathcal{F}_n$?