For the Fourier transform on $\mathbb{R}$, the answer is simple:
$$
\hat{f}(\xi)=\int_{-\infty}^\infty f(x)\,e^{-2\pi i x\cdot\xi}\,\mathrm{d}x
$$
thus, we have with the change of variables $y=ax$
$$
\begin{align}
\int_{-\infty}^\infty f(ax)e^{-2\pi i x\cdot\xi}\,\mathrm{d}x
&=\frac1{|a|}\int_{-\infty}^\infty f(y)\,e^{-2\pi i y\cdot\xi/a}\,\mathrm{d}y\\
&=\frac1{|a|}\hat{f}(\xi/a)
\end{align}
$$
However, for Fourier series on $\mathbb{T}=\mathbb{R}/\mathbb{Z}$, scaling the function only gives a nice answer if $a\in\mathbb{Z}$.
First, suppose $(a,n)=b$, then
$$
\begin{align}
\color{#C00000}{\sum_{k=0}^{a-1}e^{-2\pi ink/a}}
&=b\sum_{k=0}^{a/b-1}e^{-2\pi i(n/b)k/(a/b)}\\
&=\color{#00A000}{b\frac{e^{-2\pi in/b}-1}{e^{-2\pi in/a}-1}}\\
&=\left\{\begin{array}{cl}\color{#C00000}{a}&\color{#C00000}{\text{if }a\;|\;n}\\\color{#00A000}{0}&\color{#00A000}{\text{if }a\!\not|\;n}\end{array}\right.
\end{align}
$$
By definition,
$$
c(n)=\int_0^1f(x)\,e^{-2\pi inx}\,\mathrm{d}x
$$
therefore, for positive $a$,
$$
\begin{align}
\int_0^1f(ax)\,e^{-2\pi inx}\,\mathrm{d}x
&=\frac1{a}\int_0^af(y)\,e^{-2\pi iny/a}\,\mathrm{d}y\\
&=\frac1{a}\sum_{k=0}^{a-1}\int_0^1f(y)\,e^{-2\pi in(y+k)/a}\,\mathrm{d}y\\
&=\frac1{a}\sum_{k=0}^{a-1}e^{-2\pi ink/a}\int_0^1f(y)\,e^{-2\pi iny/a}\,\mathrm{d}y\\
&=\left\{\begin{array}{cl}c(n/a)&\text{if }a\;|\;n\\0&\text{if }a\!\not|\;n\end{array}\right.
\end{align}
$$
which, upon checking signs, works for all $a\not=0$.