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 $a\not=0$.