Suppose $Y \mid \theta \sim N(\theta, \sigma^2)$. Let $\mu_G(Y)$ be the posterior mean for $\theta$ when $\theta \sim G$ is the prior. How large can the average discrepancy in posterior mean $$ \sup_{G, G_0, \sigma} E_{Y \sim G_0 \star N(0, \sigma^2)} [(\mu_G(Y) - \mu_{G_0}(Y))^2] $$ be, where $G, G_0$ are constrained to have mean zero and variance one?
My guess is that it's bounded by the squared Wasserstein-2 distance between $G, G_0$, which is in turn bounded by 2, but I'm having trouble proving it.