We need to show that $\langle Mx,My\rangle=\langle x,y\rangle$ for all $x,y$. By definition, the change of basis matrix $M$ has the property that
$$
Mu_i=v_i
$$
Let $x=\alpha_1u_1+\ldots+\alpha_nu_n$ and $y=\beta_1u_1+\ldots\beta_nu_n$. Then, since $\{u_i\}$ is an orthonormal basis, $\langle u_i,u_j\rangle=\delta_{ij}$, the Kronecker delta. Thus, by expanding out the inner product $\langle x,y\rangle$, we see that
$$
\langle x,y\rangle=\sum_j\alpha_j\bar{\beta}_j
$$
Now compute $\langle Mx,My\rangle$:
$$
\begin{align*}
\langle Mx,My\rangle&=\langle M(\alpha_1u_1+\ldots+\alpha_nu_n),M(\beta_1u_1+\ldots+\beta_nu_n)\rangle\\
&=\langle \alpha_1Mu_1+\ldots+\alpha_nMu_n,\beta_1Mu_1+\ldots+\beta_nMu_n\rangle\\
&=\langle \alpha_1v_1+\ldots+\alpha_nv_n,\beta_1v_1+\ldots+\beta_nv_n\rangle\\
&=\sum_i\sum_j\alpha_i\bar{\beta}_j\langle v_i,v_j\rangle
\end{align*}
$$
Since $\{v_i\}$ is orthonormal, $\langle v_i,v_j\rangle=\delta_{ij}$, and so the above is exactly $\langle x,y\rangle$
The property of the dual basis is that $e^{i}(e_{j}) = \delta_{i,j}$. So, why don't you consider how $T_{U}(e_{i})$ and $e^{i}$ act on the basis of $U$?
Best Answer
Hint #1
Look at the definition of $e_j'$ and take its inner product with $e_i$. In other words, compute
$$ \langle e_i, e_j' \rangle \;\; =\;\; \left \langle e_i, \; \sum_{k=1}^n U_{kj} e_k \right \rangle. $$
Hint #2
Recall that for $U$ to be unitary we must have that $UU^* = U^*U = I$. Equivalently this can be stated that $\langle Uv, Uw\rangle = \langle u,w\rangle$ for all vectors $u$ and $w$ in your vector space. Rewrite $u$ and $w$ in the basis $\{e_1, \ldots, e_n\}$ and this should be relatively easy to demonstrate.