= Solution
In an <orthonormal basis>, every <vector> $w$ has the coordinate expansion $w=\sum_i\langle w,e_i\rangle e_i$. If $\langle Su,v\rangle=\langle u,Rv\rangle$ for all $u,v$, apply it to $u=e_i$, $v=e_j$ and take complex conjugates:
$$
\langle Re_j,e_i\rangle=\langle e_j,Se_i\rangle.
$$
Consequently
$$
\boxed{Re_j=\sum_{i=1}^n\langle e_j,Se_i\rangle e_i.}
$$
Conversely, this formula gives
$$
\langle e_i,Re_j\rangle=\overline{\langle e_j,Se_i\rangle}=\langle Se_i,e_j\rangle.
$$
Linearity in the first argument and conjugate-linearity in the second extend the equality from <basis> <vectors> to all $u,v$. This proves the equivalence.
The displayed formula specifies exactly one image for each <basis> <vector>, so it defines a unique <linear transformation> $S^*$ and proves the existence and uniqueness of the <adjoint on an inner product space>.
Conjugate symmetry gives
$$
\langle S^*u,v\rangle
=\overline{\langle v,S^*u\rangle}
=\overline{\langle Sv,u\rangle}
=\langle u,Sv\rangle.
$$
By uniqueness of the <adjoint operator>, $\boxed{(S^*)^*=S}$. Also
$$
\langle STu,v\rangle=\langle Tu,S^*v\rangle
=\langle u,T^*S^*v\rangle,
$$
so uniqueness gives $\boxed{(ST)^*=T^*S^*}$. The reversal of order follows directly from moving the two factors across the <inner product> one at a time.
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