It has been shown by Eichhorn, Linz and Hänggi in 2000 that the numerical values of Lyapunov exponents are invariant under any invertible variable transform. This is just a reformulation of the fact that they are metric invariant, because the authors presume the norm $|\cdot|$ to be an arbitrary norm in the given coordinates - just it's basic properties such as linearity are enough.
To get some intuition for this - Lyapunov exponents are linked with the Haussdorf or fractal dimension of the trajectory. Even though the Hausdorff dimension is defined on a metric space, we have an intuition that a fractal dimension is actually more of a property of differential structure rather than of a specific notion of length/surface/volume. The metric is just a handle to get to the fractal dimension, but it's nature is non-metric. We can understand acquiring Lyapunov exponents in a similar way - the metric is just a handle and we choose one arbitrarily.
A second way to get an intuition is through the explicit definition of the exponents $\lambda$ via the linear variation $\delta x(t)$ evolved in time:
$$\lambda = \lim_{t\to \infty} \frac{\log |\delta x(t)|}{t}$$
Let us assume that $\delta x(t) = e^{\mu t}\delta x(0)$. Then out of the linearity of the norm we have $|\delta x(t)| = e^{\mu t} |\delta x(0)|$ and the limit yields
$$\lambda = \lim_{t \to \infty} (\frac{\mu t}{t} + \frac{\log |\delta x(0)|}{t})$$
The second term dies off and we have $\lambda=\mu$ for any positive definite linear $|\cdot|$. I.e. you get the same number with a different norm and thus the Lyapunov exponent gives you something which is connected to a "relative growth rate" independent of the metric. There are some loopholes to this argument and these are covered by the article cited above.