 Solution

A linear code of length $n$ and rank $k$ is a $k$-dimensional linear subspace of ${\mathbb{F}}_2^n$. The Hamming weight of a vector counts its nonzero coordinates. Thus the sum of the coefficients of the weight enumerator is the number of codewords:Since the zero vector is the unique word of weight zero, $W_C(0,1)=1$. Symmetry of the enumerator therefore implies $W_C(1,0)=1$. Conversely $W_C(1,0)=1$ means that the all-one vector $1$ belongs to $C$. Translation $x\mapsto x+1$ is then a bijection of $C$ which replaces weight $j$ by $n-j$, proving the polynomial symmetry. For length zero the same conclusion holds with the unique empty vector.

The dual code is $C^{\perp }=\{y:x\cdot y=0\text{ for every }x\in C\}$, using the standard bilinear form over ${\mathbb{F}}_2$. If $y\in C^{\perp }$, each summand of the character sum equals one. Otherwise choose $x_0\in C$ with $x_0\cdot y=1$; replacing $x$ by $x+x_0$ permutes the summands and reverses their sign. The sum equals its own negative and must be zero. ConsequentlyThis is character orthogonality for the additive group of the code.