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Use tangent vectors, rather than the affine tangent plane, to differentiate. Every skew-symmetric matrix $A$ gives a curve $X\exp (tA)$ in the special orthogonal group. Its first derivative under $f$ is $\mathrm{tr}(CXA)$. The trace pairing annihilates every skew-symmetric matrix precisely when $CX$ is symmetric: choosing $A$ with only entries $a_{ij}=-a_{ji}$ tests each off-diagonal antisymmetric entry. Thus criticality is equivalent to $Y=CX$ being symmetric.

Since $X$ is orthogonal, $Y{Y}^T=C^2$, so at a critical point $Y^2=C^2$. Consequently $Y$ commutes with $C^2$. The diagonal entries $c_i^2$ are distinct, and the equation $Y{C}^2=C^{2}Y$ gives $(c_j^2-c_i^2)y_{ij}=0$ for $i\ne j$. Hence $Y$ and $X$ are diagonal. Orthogonality and the determinant constraint give exactlyConversely each such $D$ makes $CD$ symmetric, so this list is complete.

To calculate the Hessian matrix, use local coordinates $X=\exp (A)D$ with $A^T=-A$; the matrix exponential is a local coordinate map near zero. Its quadratic term givesbecause $(A^2{)}_{ii}=-{\sum }_{j\ne i}{a}_{ij}^2$. There are no mixed quadratic terms. For $i<j$, $c_j>c_i>0$, so $c_i{\epsilon }_i+c_j{\epsilon }_j$ has the sign of ${\epsilon }_j$ and is never zero. Every listed critical point is therefore nondegenerate, with Morse indexThis is the trace height on the special orthogonal group calculation; the rotations in the hint select the individual coordinates $a_{ij}$.

For $n=3$, the signs $(+--),(-+-),(--+),(+++)$ have indices $0,1,2,3$ respectively. A Morse function on a compact closed manifold has Euler characteristic equal to the alternating number of critical points, by its Morse handle-attachment theorem and resulting finite CW complex. Hence