Not sure whether the claim is true or false.
If claim is true, intuitively, it might have something to do with "least favorable priors", but am not able to figure out the connection.
If claim is false, one example is when $X_i|\theta \sim $ Poisson$(\theta)$, then $\bar{X}$ is minimax. But a Gamma$(\alpha, \beta)$ prior fails since, that would indicate $\beta = 0$, which is improper. But, how do we know there is no other prior that gives $\bar{X}$ as a Bayes rule?
So I think I figured out a valid counter example.
$X_i \overset{iid}\sim N(\theta, 1)$. Then $\bar{X}$ is minimax. (This can be shown by using a sequence of priors). However, $\bar{X}$ being unbiased for $\theta$, can only be Bayes for any proper prior if the Bayes Risk is 0.
$R(\theta,\bar{X}) =\dfrac{1}{n} \implies \int_{\Theta} R(\theta,\bar{X}) \pi(\theta) d\theta = \dfrac{1}{n} \int_{\Theta} \pi(\theta) d\theta = \dfrac{1}{n}$
Thus, as long $\pi$ is proper, Bayes risk of $\bar{X}$ cannot be 0. Hence $\bar{X}$ cannot be Bayes with respect to any proper. The same reasoning can be made for admissible estimators.
