= Solution
Use <normalized Fourier analysis on a finite abelian group>, identifying $A$ with its <indicator function> $a=\mathbf1_A$ on the <cyclic group> $\mathbb Z_n$. The conventions are
$$
\mathbb E_xh(x)=\frac1n\sum_{x\in\mathbb Z_n}h(x),\qquad
\widehat h(r)=\mathbb E_xh(x)\omega^{-rx},\qquad
(h*k)(x)=\mathbb E_yh(y)k(x-y).
$$
The last operation is <normalized convolution on a finite group>. With these conventions the <Fourier coefficients on a finite abelian group> use a normalized average, while sums over frequencies use counting measure.
Expanding the <normalized convolution on a finite group> and putting $z=x-y$ gives
$$
\widehat f(r)=\mathbb E_x\mathbb E_y a(y)a(x-y)\omega^{-rx}
=\bigl(\mathbb E_y a(y)\omega^{-ry}\bigr)\bigl(\mathbb E_z a(z)\omega^{-rz}\bigr)
=\widehat a(r)^2.
$$
For the <translation of a function> $g(x)=f(x-u)$, putting $z=x-u$ yields
$$
\widehat g(r)=\mathbb E_z f(z)\omega^{-r(z+u)}=\omega^{-ru}\widehat f(r).
$$
\b[The two transforms are therefore]
$$
\boxed{\widehat f(r)=\widehat A(r)^2,\qquad \widehat g(r)=\omega^{-ru}\widehat A(r)^2.}
$$
The negative sign in the translation factor follows from the negative sign in our <Fourier coefficient on a finite abelian group> convention.
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