Why is the equation $E[X_s|\mathcal{F_s}] = X_s$? I understand the algebraic manipulations that get $E[X_t|\mathcal{F_s}] = X_s$ but can't build intuition on what the point of having a conditional expectation - given the filtration - is if it's just gonna be that it's constant.
An application of this fact would be useful. If, $E[X_s|\mathcal{F_s}] = X_s$ then is the $X_s$ on the RHS still a random variable?
It is important to remember that the conditional expectation is a random variable (hence not constant) that acts as the projection on presupposed information. So in your case $E[X_t|\mathcal{F}_s]$ is the 'projection' of $X_t$ as an $L^1$-function onto the subspace of $L^1$-functions that is 'known' up to time $s$ - the closest random variable to $X_t$ whose values are known from $\mathcal{F}_s$. To say that $E[X_s|\mathcal{F}_s] = X_s$ is simply to say that we already know all that is to know from $X_s$ at time $s$ so any added information at time $s$ (that is $\mathcal{F}_s$) is redundant: the projection of an element on the projected subspace is identical. Thus your intuition is correct in some sense that it is trivial (by adaptedness), but that the martingle-property $E[X_t|\mathcal{F}_s] = X_s$ is highly non-trivial in the sense of projections. Kallenberg's Foundations of Modern Probability treats conditional expectations in this way.
It is also important to realize here the dependence of martingales and its filtration, that is that projecting a random variable onto any other 'known information' may inhibit the martingale property.