= Solution
The <gradient> vanishes at $0$. Decompose any <vector> as $x=tb+w$, with $w\perp b$. <Symmetry> of $A$ and $Ab=\tfrac12b$ imply $b^TAw=(Ab)^Tw=0$. Since $\|b\|=1$,
$$
g(tb+w)=-\tfrac12t^2+w^TAw.
$$
If $d\geq2$, choose a nonzero $w\perp b$. Arbitrarily close to zero, the direction $b$ gives negative values and the direction $w$ gives positive values. Thus \b[$0$ is a <saddle point> for $d\geq2$]. Equivalently, the <Hessian matrix> has one negative <eigenvalue> and all its other <eigenvalues> positive.
The dimension matters: if $d=1$, the <orthogonal complement> is zero-dimensional and $g(tb)=-t^2/2$. In that case \b[$0$ is the unique <global maximum>]. This edge case is not excluded by the stated hypotheses. The decomposition is a <rank-one downdate along an eigenvector>.
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