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A Converse of the Skorokhod Embedding Theorem

I am wondering whether the following "sort of converse" of Skorokhod's embedding theorem holds:

Suppose that $\{D_t\}_{t \geq 0}$ is a stochastic process with continuous paths, $D_0 = 0$, and suppose that the following is true:

For any mean zero, finite variance random variable $X$ there exists a stopping time $\tau$ with respect to the filtration $\mathcal{F}_t \stackrel{def}{=} \sigma\{D_s:0\leq s\leq t\}$, such that $X \stackrel{d}{=} D_\tau$ and $\mathbb{E}X^2 = \mathbb{E}\tau$.

My question is, does this imply that the process $\{D_t\}$ is a Brownian motion?

Even if not, under what condition(s) does it follow that $\{D_t\}$ is a Brownian motion? For example does strong Markov property or independent increment assumption of $\{D_t\}$ help?

Any help will be greatly appreciated.