I'm not an expert, but I would say that the function $f(x) = a x^k$ is scale invariant in the following sense:
Consider some interval $(x, b x)$. In this interval, the function $f(x)$ goes from:
\begin{aligned}
f(x) \quad &\text{ at }x,\\
f(bx) = b^{k}f(x) \quad &\text{ at }bx.
\end{aligned}
Now, imagine that you scale the interval by some value $\lambda$. In other words, let's look at the interval $(\lambda x, \lambda b x)$. If we do the same analysis as before, we can see that the function $f(x)$ over the interval goes from
\begin{aligned}
f(\lambda x) = \lambda^{k} f(x) \quad &\text{ at }\lambda x,\\
f(\lambda bx) = \lambda^{k}b^{k}f(x) \quad &\text{ at }\lambda bx.
\end{aligned}
So you should be able to see that over the "scaled" interval, the function $f(x)$ behaves exactly like it does over the "unscaled" interval, except for a global multiplicative factor.
What this means is that if you look at a function like $f(x)$ over a range of (appropriately scaled) intervals, it will look identical, modulo the above-mentioned factor. If you'd like to see this more clearly, I've plotted the parabola that you mention below on Mathematica, over the range $(\lambda , \,2\lambda)$, while varying $\lambda$ from 1 to 100. You should be able to see that the form of the function remains the same (the "jiggling" you see is because the successive images are not aligned well):
Manipulate[Plot[x^2, {x, \[Lambda], 2 \[Lambda]}, PlotRange -> {{\[Lambda], 2 \[Lambda] }, {\[Lambda]^2, 4 \[Lambda]^2}}, PlotStyle -> {Black}], {\[Lambda], 1, 100}]
On the other hand, if you did the same thing with a function that isn't scale invariant like the exponential (as @GiorgioP suggested in a comment below), you'd see something very different for the same parameter range:
