Given a stochastic process $\{X_{t}\}_{t\ge 0}$ and a filtration $\{\mathcal{H}_{t}\}_{t \ge 0}$ generated by $\{X_{t}\}_{t\ge 0}$ and if $X_{t}$ is a martingale w.r.t. $\mathcal{H}_t$ we have the property
$1)$ $E[X_{t}] =E[X_0]$ for all $t \ge 0$
the proof to this goes as follows using the property $E[X_{t}|\mathcal{H}_{s}]=X_{s}$ for $0 \le s \le t$ we have
$E[X_{t}|\mathcal{H_{0}}]=X_0$
Then taking the expectation we get the desired result $(1)$.
My question is this what is stopping us doing the same for $\mathcal{H}_1$ and then deducing via the same method that $E[X_t] =E[X_1]$ for all $t \ge 1$ surely this contradicts $(1)$?
There's no contradiction. In fact, (1) implies that $E(X_t) = E(X_1)$ for all $t \geq 1$.