A particular commutator of the discrete Fourier matrix For $N$ be a fixed natural number, define $w=e^{\frac{2\pi i}{N}}$ and $z=e^{\frac{\pi i}{N}}$, so that $z^2=w$. Let $D$ be the diagonal matrix $D=\operatorname{diag}(1,z,z^2,\ldots,z^{N-1})$ and $F$ be the Discrete Fourier matrix of order $N$, that is $F=\frac{1}{\sqrt{N}}(w^{kl})_{0\leq k,l\leq N-1}$.

Q. Does there exist any   invertible matrix  $A$ such that the following identity
$$A(DF)A^{-1}=\alpha_A DF$$
holds for some complex number $\alpha_A\neq1$.

 A: If such a pair $(A,\alpha\ne1)$ existed, then the spectrum of $DF$ would be invariant under the operation $\lambda\mapsto\alpha\lambda$, in particular $\alpha$ would be a root of unity.
The answer is therefore negative as soon as $N=2$, for then
$$D=\begin{pmatrix} 1 & 0 \\ 0 & i \end{pmatrix},\qquad
F=\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix},\qquad DF=\begin{pmatrix} 1 & 1 \\ i & -i \end{pmatrix}.$$
This matrix has distinct eigenvalues
$$\lambda_\pm=\frac{1-i}2\left(1\pm i\sqrt{\frac72}\right).$$
However the ratio
$$\alpha:=\frac{\lambda_+}{\lambda_-}=\frac{\sqrt2+i\sqrt7}{\sqrt2-i\sqrt7}=\frac{-5+2i\sqrt{14}}9$$
is not a root of unity.
Edit. This is a general analysis. If $M:=DF$ is such as in the question, then $M$ is conjugated to $\alpha M$. Since $M$ is unitary (as the product of unitary matrices) its spectrum is made of non-zero numbers (actually of unit moduli). In particular $\alpha$ is a root of unity. The eigenvalues can be split into classes (taking in account their mutilpicities), which shows that the order $r\ge2$ of $\alpha$ divides $N$. All this shows that the characteristic polynomial $\chi_M(X)$ actually depends only upon $X^r$.
Conversely, if $\chi_M(X)=Q(X^r)$, then the spectrum is invariant under $\lambda\to \alpha\lambda$ with $\alpha^r=1$, hence $M$ is conjugated to $\alpha M$ (use the fact that $M$ is diagonalisable).
Thus a necessary and sufficient condition is that $\chi_M$ writes as $Q(X^r)$ for some $r\ge2$ dividing $N$.
In particular a necessary condition is that
$${\rm Tr}M\quad\left(=\sum_{k=0}^{N-1}z^{2k^2+k}\right),$$
the coefficient of $-X^{N-1}$, vanishes. This turns out to be false for every $N\le7$ and probably most of the time.
Therefore, I claim that the answer is negative for $N\le7$, and probably most of the time.
Calculation trick : because $M^{-1}=\bar F\bar D$ is conjugated to $\bar M$, we have
$$\bar\chi_M(X)=\det(-M)^{-1}X^N\chi_M(X^{-1}).$$
Thus we need only to check whether the first half of the polynomial $\chi_M$ writes as a polynomial in $X^r$.
Follow-up. François Brunault's answer to a Q of mine sets up that the trace of $M$ never vanishes. Thus the answer to OP's question is always negative.
