This is part of the fifth exercise of the book An introduction to harmonic analysis of Yitzhak Katznelson.
Show that if $m\nmid n$ then $\widehat{f_{(m)}}(n)=0$
Here it is assumed that $f\in L_1(\Bbb R/\Bbb Z,\Bbb C)$
We have that $$ \begin{align} \widehat{f_{(m)}}(n)&=\frac1{2\pi}\int_0^{2\pi}f(mt)e^{-int}dt \\&=\frac1{2\pi m}\int_0^{2\pi m}f(t)e^{-int/m}dt\\ &=\frac1{2\pi m}\sum_{k=0}^{m-1}\int_{2\pi k}^{2\pi (k+1)}f(t)e^{-int/m}dt\\ &=\frac1{2\pi m}\sum_{k=0}^{m-1}\int_{0}^{2\pi }f(t)e^{-i\frac nm(t+2\pi k)}dt\\ &=\frac1{2\pi m}\int_{0}^{2\pi }f(t)e^{-i nt/m}\sum_{k=0}^{m-1}e^{-i2\pi kn/m}dt \end{align} $$
Then if $n/m=p+r/m$, for some $p,r\in\Bbb N$ where $r<m$, then everything reduces to show that
$$\sum_{k=0}^{m-1}e^{-i2\pi kn/m}=\sum_{k=0}^{m-1}e^{-i2\pi kr/m}=0$$
for any $r\in\{1,2,\ldots,m-1\}$. The case of $r=1$ is trivial because the exponentials are the $m$-th roots of unity, the solutions to $x^m=1$, and is clear that $[x^{m-1}](x^m-1)=\sum_{k=0}^{m-1}e^{-i2\pi k/m}=0$.
However Im stuck for the other cases. I need some help.
Proof assuming that $f$ is a continuous periodic function. If $f(x)=e^{ikx}$ then $\hat f_{(m)}(n)=\frac 1 {2\pi} \int_0^{2\pi} e^{i(kmx-nx)}dx=0$ because $km \neq n$. Hence $\hat f_{(m)}(n)=0$ for any trigonometric polynomial $f$. Since $f$ is a uniform limit of trigonometric polynomials we are done.
Proof for $L^{2}$ periodic functions is identical since we can again approximate $f$ in $L^{2}$ by trigonometric polynomials.
More generally the result is true for $L^{1}$ periodic functions also since you can approximate them in $L^{2}$ norm by $L^{2}$ periodic functions.