It is easy to see that the the floor function on $[0,1]$ has Fourier expansion: $$[x] = x - {1 \over 2} + {1 \over \pi }\sum\limits_{n > 0} {{{\sin (2\pi x)} \over n}}.$$
Let $a$ be a positive integer. My question is: what about the function $[ax]$, can I replace $x$ by $ax$ in the Fourier series? Thanks.
Short answer: Fourier expansions depend on the period of the function (or the period we impose on the function, if it's defined only on an interval).
First of all, the equation $$ [x] = x - {1 \over 2} + {1 \over \pi }\sum\limits_{n > 0} {{{\sin (2\pi nx)} \over n}} $$ is true for all real numbers $x$ (except integers, where the left- and right-hand sides account differently for the jump discontinuity). However, this is not technically a Fourier expansion of $[x]$ on $[0,1]$; as commenters have pointed out, that function is $0$ almost everywhere and thus has the trivial Fourier expansion.
What is really intended here is that the function $x-[x]$ is periodic with period $1$, and therefore has a well-defined period-$1$ Fourier expansion, which is indeed ${1 \over 2} - {1 \over \pi }\sum_{n > 0} {{{\sin (2\pi nx)} \over n}}$.
Now, it is a true statement that $$ [ax] = ax - {1 \over 2} + {1 \over \pi }\sum\limits_{n > 0} {{{\sin (2\pi nax)} \over n}} $$ (just substitute $ax$ for $x$). This is the correct period-$\frac1a$ Fourier expansion of the periodic function $ax-[ax]$.
However, if for some reason you wanted to define a function $f(x)$ to equal $ax-[ax]$ for $0\le x<1$, and then extend the function $f(x)$ to be periodic with period $1$, then its Fourier expansion would be (for generic values of $a$) very different than the formula above.