Given a basis $a_\nu$ of a vector space $E$, define the mapping $\phi: E\to E$ as $$\phi (a_\nu) = \sum_\mu a_\mu.$$ Find the matrix of the dual mapping relative to the dual basis.
Since $M(\phi) = (\alpha_\nu^\mu)$, where $\alpha_\nu^\mu = 1$, and the matrix of the dual mapping relative to the dual basis has to the property $$M(\phi)^\perp = M(\phi^*),$$ where $\phi^*$ is the dual mapping of $\phi$ relative to the dual basis of $a_\nu$, and $\perp$ means the transpose of the matrix.
So, this directly says that $M(\phi^*) = (\alpha_v^\mu)$, where $\alpha_\nu^\mu = 1.$
I don't see any problem up to this point, but this is a question that is asked in Graduate Linear Algebra book, and anyone who has read the book can give an answer to this question, so this suggested that I'm missing something in the question because there is not point in asking such a obvious question, so is my understanding and my answer are correct? If not, what am I missing?
Your understanding is correct, once you know that the required matrix is the transpose.
If you want to prove it directly, you need to express $\phi^*(a_\nu^*)$ in terms of the basis. Now $$ \phi^*(a_\nu^*)(a_\lambda) =a_\nu^*\circ\phi(a_\lambda) =a_\nu^*\Bigl(\sum_\mu a_\mu\Bigr) =1 $$ so, for $x=\sum_\mu c_\mu a_\mu\in E$, $$ \phi^*(a_\nu^*)(x) =\phi^*(a_\nu^*)\Bigl(\sum_\mu c_\mu a_\mu\Bigr) =\sum_\mu c_\mu =\sum_\mu a_\mu^*(x) $$ and therefore $$ \phi^*(a_\nu^*)=\sum_\mu a_\mu^* $$