Given $X(t) = t W(1/t)$, I want to show that $X(t)$ is a Brownian Motion. Specifically, I want to show it using Levy's Characterization of Brownian Motion. Levy's Characterization of Brownian Motion states that if a process is almost surely a continuous martingale with $X(0) = 0$ and quadratic variation $t$, then the process is a Brownian Motion. I know how to show the continuity of $X(t)$ and that the quadratic variation is equal to $t$, but it is not clear to me why this process is a martingale. In my attempt to show this I have done the following:
\begin{align} E(X(t) | \mathcal{F}_s) &= E(t W(1/t) | \mathcal{F}_s)\\\\ &= tE(W(1/t) | \mathcal{F}_s)\\\\ &= t W(1/s). \end{align} My understanding is that the martingale property states that $E(X(t) | \mathcal{F}_s) = X(s)$, so, in this case, we would need $E(tW(1/t) | \mathcal{F}_s) = sW(1/s)$ which we do not have. Is this correct or am I mistaken?