= Solution
The <adjoint operator> $T^*$ is characterized by
$$
\boxed{\langle Tv,w\rangle=\langle v,T^*w\rangle\quad\text{for all }v,w\in V.}
$$
This identity works with either convention for which argument of the complex <inner product> is linear, provided the convention is used consistently.
For uniqueness, subtract the two given identities to get $\langle v,(S-T^*)w\rangle=0$ for every $v,w$. Fix $w$ and choose $v=(S-T^*)w$. Then <positive definiteness> gives
$$
\|(S-T^*)w\|^2=0,
$$
so $(S-T^*)w=0$. As $w$ was arbitrary, $\boxed{S=T^*}$. The decisive fact is that a vector orthogonal to every vector, including itself, must be zero.
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