The tag ergodic-theory looks inappropriate. There is no measure-preserving map related to your question.

I agree with that, provided $B$ independent of $X_0$. You use that twice, and you should mention it. That $B$ is independent of $X_0$ is not so obvious, although it looks true.

@Tony419 It was clear for me that $\phi+\psi$ was $\mathcal{C}^1$ (by looking at left and right derivatives at $0$), and that its derivative has finite variation. When I derivated once more, I was surprised to see that $\phi+\psi$ is $\mathcal{C}^2$. There is no deeper philosophy.

I would agree with this comment in the present situation if $\hat{\phi}$ was assumed to be in the Schwarz space.

The simple way is to assume that $X$ is right-continuous, that $c$ is continuous at $x$ and bounded. Then Lebesgue dominated convergence theorem applies after a change of variable $s=tu$.

I do not see in the assumptions what prevents the random variables $X_1,X_2,\ldots$ from being equal.

@Nate River. Thank you. I still wonder whether your conclusion holds if you assume that that all $\mathcal{F}$-martingales are still $\mathcal{FH$-martingales. The counterexample I gave does not answer this alternative question.

Ah yes, you assume that $\mathcal{F}$ is the natural filtration of $M$. Anyway, the answer is negative, see the answer below.

I you wish the condition to be necessary (I think it it not, but for the moment, I have no counterexample), you should at least assume that ALL $\mathcal{F}$-martingales are still $\mathcal{H}$-martingales. This property is called immersion of filtrations. Otherwise, the null martingale provides trivial counterexamples.