Let $(M_t)$ is a martingale for $t \in [0,1]$.
Then, we immediately have from the martingale property that $E[M_0]=E[M_t]$ for all $t \in [0,1]$ and $E[M_s^2]\leq E[M_t^2]$ for all $s\leq t$ in $[0,1]$.
When do we have $E[M_s^2]= E[M_t^2]$ for all $s\leq t$ in $[0,1]$ ? I feel that $(M_t)$ must be constant a.s.. Are there martingales that are not constants that satisfy $E[M_s^2]= E[M_t^2]$ for all $s\leq t$ in $[0,1]$ ?
For $s\leqslant t$, denoting by $\left(\mathcal F_t\right)$ the filtration for which $(M_t)$ is a martingale, we have $$ \mathbb E\left[M_t^2\mid\mathcal F_s\right]=\mathbb E\left[\left(M_t-M_s+M_s\right)^2\mid\mathcal F_s\right]=\mathbb E\left[\left(M_t-M_s\right)^2\mid\mathcal F_s\right]+\mathbb E\left[M_s^2\mid\mathcal F_s\right] +2\mathbb E\left[\left(M_t-M_s\right)M_s\mid\mathcal F_s\right] $$ hence the martingale property gives $$ \mathbb E\left[M_t^2\mid\mathcal F_s\right] =\mathbb E\left[\left(M_t-M_s\right)^2\mid\mathcal F_s\right]+ M_s^2 . $$ As a consequence, $$ \mathbb E\left[M_t^2 \right] =\mathbb E\left[\left(M_t-M_s\right)^2 \right]+ \mathbb E\left[M_s^2\right] . $$ Therefore, we should have $\mathbb E\left[\left(M_t-M_s\right)^2 \right]=0$ for each $s\leqslant t$.