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 particular commutator of the discrete Fourier matrix
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](https://mathoverflow.net/q/420496) to a Q of mine sets up that the trace of $M$ never vanishes. Thus the answer to OP's question is **always negative**.


