= Solution
Use the <Dirac notation> convention that the <inner product> is linear in the ket. Orthonormality of the computational <basis> gives
$$
\langle k|f\rangle=\sum_{\ell\in\mathbb Z_N}f(\ell)\langle k|\ell\rangle=f(k).
$$
The finite geometric sum satisfies $\sum_{k=0}^{N-1}\lambda^{rk}=N$ if $r=0$ in $\mathbb Z_N$, and zero otherwise: in the second case it equals $(1-\lambda^{rN})/(1-\lambda^r)=0$. Therefore
$$
\langle\widehat m|\widehat n\rangle=\frac1N\sum_k\lambda^{(n-m)k}=\delta_{mn}.
$$
There are $N$ such orthonormal <vectors> in an $N$-dimensional space, so they form a <basis>. The inverse expansion is verified by the same sum:
$$
\boxed{\frac1{\sqrt N}\sum_j\lambda^{-jk}|\widehat j\rangle
=\frac1N\sum_{j,\ell}\lambda^{j(\ell-k)}|\ell\rangle=|k\rangle.}
$$
This is the <unitary> <discrete Fourier transform> and its inverse.
For the <basis-copying isometry>, the images $|k\rangle|k\rangle$ are orthonormal, so $C$ is a linear <isometry>, with $C^*C=I$. Its <matrix> elements give
$$
\langle Ck|\,|f\rangle|g\rangle
=\langle k|f\rangle\langle k|g\rangle=f(k)g(k)=\langle k|f.g\rangle.
$$
Applying the definition of the <adjoint operator>, the coefficient of $|k\rangle$ in $C^*|f\rangle|g\rangle$ is exactly this expression. Thus
$$
\boxed{C^*|f\rangle|g\rangle=|f.g\rangle.}
$$
Equivalently $C^*|k\rangle|\ell\rangle=\delta_{k\ell}|k\rangle$. The adjoint of <basis> copying acts as pointwise multiplication of coefficient <functions>.
For the <Fourier transform of a basis-copying isometry>, the <normalization> matters:
$$
f_m(k)f_n(k)=\frac{\lambda^{(m+n)k}}N=\frac1{\sqrt N}f_{m+n}(k).
$$
Hence $C^*|\widehat m\rangle|\widehat n\rangle=N^{-1/2}|\widehat{m+n}\rangle$. Taking adjoint <matrix> elements, or expanding both output factors in the inverse Fourier <basis>, yields
$$
\boxed{C|\widehat j\rangle=\frac1{\sqrt N}\sum_{m+n=j}|\widehat m\rangle|\widehat n\rangle.}
$$
\b[The printed formula is missing the factor $1/\sqrt N$.] There are $N$ mutually <orthogonal> terms in its sum, so without that factor its right side has <norm> $\sqrt N$, whereas its left side has <norm> one because $C$ is an <isometry>. The formula above is the normalized identity consistent with all the preceding definitions. For $N=1$ the distinction disappears, but the no-cloning assertion is substantive for $N\geq2$.
The normalized output is not $|\widehat j\rangle|\widehat j\rangle$. Its coefficient <matrix> pairs every Fourier label $m$ with $j-m$; tracing out either factor gives $I/N$. Thus it is a <maximally entangled state>, while two clones of a pure Fourier <quantum state> would have pure reduced <quantum states>. A copier for the computational <basis> therefore does not copy the Fourier <basis>.
More generally, suppose a linear <isometry> with a fixed blank input could clone every normalized <quantum state>. Its restriction to $|\psi\rangle|\mathrm{blank}\rangle$ would define a linear map $D$ with $D|k\rangle=|k\rangle|k\rangle$ on every computational <basis> <quantum state>. Linearity forces $D=C$, which has just been shown to fail on Fourier <quantum states>. This proves the <no-cloning theorem> in the pure-<quantum state> isometric setting.
The inner-product proof makes the obstruction independent of the chosen <basis>. If an <isometry> cloned two <quantum states> with overlap $s=\langle\psi|\phi\rangle$, preservation of inner products would give $s=s^2$, or, allowing irrelevant output phases, $|s|=|s|^2$. Distinct nonorthogonal rays have $0<|s|<1$, which is impossible. Even allowing an output environment does not help: <unitary> evolution would give $|s|=|s|^2|\langle e_\psi|e_\phi\rangle|\leq|s|^2$, the same contradiction. \b[Arbitrary unknown <quantum states> cannot be cloned], although any fixed <orthonormal basis> can be copied and a one-dimensional <quantum state> space is trivial.
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