= Solution
Here $n$ is a positive <integer> and $m$ is an <integer> index of a <Fourier coefficient>. Split a period into $n$ intervals and set $y=nx-2\pi j$ on the $j$th interval. Periodicity and integrability give
$$
\begin{aligned}
\int_0^{2\pi}f(nx)e^{imx}\,dx
&=\frac1n\sum_{j=0}^{n-1}e^{2\pi imj/n}
\int_0^{2\pi}f(y)e^{imy/n}\,dy.
\end{aligned}
$$
For $0<|m|<n$, $\zeta=e^{2\pi im/n}$ is a <root of unity> different from one and satisfies $\zeta^n=1$, so $\sum_{j=0}^{n-1}\zeta^j=(1-\zeta^n)/(1-\zeta)=0$. Therefore
$$
\boxed{\int_{\mathbb T}f(nx)e^{imx}\,dx=0\qquad(0<|m|<n).}
$$
For $m=0$ the same substitution gives $\int_{\mathbb T}f(nx)\,dx=\int_{\mathbb T}f(x)\,dx$. If the <mean> of $f$ is zero, this last mode vanishes as well. Since the <linear span> of the exponentials with $|m|\leq n-1$ is the space of <trigonometric polynomials> of degree at most $n-1$, every pairing with that space is zero. This is the <periodic dilation annihilates low Fourier modes> property. The integer-index assumption is essential; the stated Fourier orthogonality would generally be false for noninteger $m$.
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