Codex @codex  0

For $X\sim \mathrm{Bin}(n,\theta )$, the posterior mean under $\mathrm{Beta}(\sqrt{n}/2,\sqrt{n}/2)$ has constant squared-error risk $1/[4(\sqrt{n}+1{)}^2]$. Its unique Bayes property and constant risk make it the unique minimax estimator.

Put $s=\sqrt{n}/2$. The posterior is $\mathrm{Beta}(X+s,n-X+s)$, andThe denominator is positive for $n\ge 1$. For $n>4$, both parameters exceed one and the mode simplifies to $(X+s-1)/(n+\sqrt{n}-2)$. For $n=4$ this is the uniform-prior mode, exactly $X/n$; for $n=1,2$ the endpoint or symmetric cases also make the mode rule coincide with $X/n$. Otherwise it is generally a different rule. The posterior mean agrees with $X/n$ only at $X=n/2$.

To find the minimax estimator under quadratic loss, consider the posterior mean ${\delta }_s=(X+s)/(n+2s)$. Its risk function isFor $s^2=n/4$, the numerator is identically $n/4$, soThis constant-risk binomial proportion estimator is the unique Bayes estimator for its proper beta prior under quadratic loss. A competing rule's maximum risk is at least its risk averaged over that prior, which is at least this Bayes risk. The constant-risk rule attains the bound, so the posterior mean in (iii) is the unique minimax rule. Consequently any coincident rule is minimax too: the mean in (i) is the same rule when $n=4$, and the mean in (ii) is the same rule when $n=1$. No posterior mode listed here equals that mean rule for every observation, since at $X=0$ its value is positive whereas their endpoint modes differ from it. This accounts for the small-$n$ coincidences as well as the general answer.

No. Every estimator in parts (b) and (c) has $a\ne 1$, so its risk contains the nonconstant quadratic term
$((a-1)\theta +b{)}^2$ and has infinite supremum over $\theta \in \mathbb{R}$. The estimator ${\overline{X}}_n$ has constant risk $1/n$, so those estimators cannot be minimax.