= Solution
Put $s=\sqrt n/2$. The posterior is $\operatorname{Beta}(X+s,n-X+s)$, and
$$
\boxed{\mathbb E[\theta\mid X]=\frac{X+\sqrt n/2}{n+\sqrt n},\qquad m_{\theta\mid X}=\frac{\max(X+s-1,0)}{\max(X+s,1)+\max(n-X+s,1)-2}.}
$$
The denominator is positive for $n\geq1$. 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> is
$$
R(\theta,\delta_s)=\frac{n\theta(1-\theta)+s^2(1-2\theta)^2}{(n+2s)^2}.
$$
For $s^2=n/4$, the numerator is identically $n/4$, so
$$
\boxed{R(\theta,\delta_s)=\frac1{4(\sqrt n+1)^2}\quad\text{for all }\theta.}
$$
This <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 \b[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.
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