= Solution
The <matrix exponential> is the absolutely convergent series $e^{tA}=\sum_{n=0}^\infty t^nA^n/n!$, with derivative $Ae^{tA}$ and inverse $e^{-tA}$. If $K^T=-K$, then $(e^{tK})^Te^{tK}=e^{-tK}e^{tK}=I$. Its <determinant> is <continuous> in $t$, is never zero, and starts at one, so it is one throughout: $e^{tK}\in SO(3)$. Conversely, differentiating $(e^{tK})^Te^{tK}=I$ at zero gives $K^T+K=0$. Thus
$$
\boxed{e^{tK}\in SO(3)\text{ for all real }t\iff K^T=-K.}
$$
The quantifier matters: membership at one isolated time alone would not imply skewness.
Write $P(F)=DW(F)$ and $J=\det F>0$. For any skew matrix $K$, differentiate frame indifference along $e^{tK}F$:
$$
0=P(F):(KF)=(P(F)F^T):K.
$$
A matrix orthogonal to every skew matrix is symmetric. Therefore $P(F)F^T$ and the <Cauchy stress tensor> $T=J^{-1}P(F)F^T$ are symmetric.
Let $\mathcal C=D^2W(I)$ and assume $P(I)=0$. Differentiating symmetry of $P(F)F^T$ at $I$ in direction $H$ shows $\mathcal C[H]$ is symmetric for every $H$. Hence $c_{ijkl}=c_{jikl}$. As a <Hessian>, $\mathcal C$ also has pair-interchange symmetry $c_{ijkl}=c_{klij}$. Combining the two gives $c_{ijkl}=c_{ijlk}$. Equivalently $\mathcal C$ annihilates <skew-symmetric matrices>[infinitesimal rotations]. This proves both requested minor symmetries and shows exactly where the stress-free assumption enters. It is <stress-free linearization of an objective energy>.
For the particular energy, $\sum_iv_i^2=\operatorname{tr}(F^TF)$ and $\prod_iv_i=J$, so
$$
P(F)=4\alpha F+h'(J)\operatorname{cof}F.
$$
At $I$ this is $[4\alpha+h'(1)]I$. Therefore the reference is stress-free exactly when $\boxed{h'(1)=-4\alpha}$. To differentiate again, use $\operatorname{cof}F=JF^{-T}$. The <determinant> and inverse derivatives at $I$ give
$$
\boxed{D(\operatorname{cof})(I)[H]=(\operatorname{tr}H)I-H^T,}
$$
which also proves the supplied hint. It follows that
$$
\mathcal C[H]=4\alpha H+h''(1)(\operatorname{tr}H)I-4\alpha[(\operatorname{tr}H)I-H^T].
$$
For a symmetric $e$ this becomes
$$
\boxed{\mathcal Ce=8\alpha e+[h''(1)-4\alpha](\operatorname{tr}e)I=2\mu e+\lambda(\operatorname{tr}e)I,\quad\mu=4\alpha,\ \lambda=h''(1)-4\alpha.}
$$
No additional positivity of the <bulk modulus> follows from the stated smoothness of $h$ alone; this calculation identifies the moduli rather than imposing extra material assumptions.
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