Consider $\mathbb{R}^n$ as a differentiable manifold with the standard coordinates $x^j: \mathbb{R}^{n} \to \mathbb{R}, x_1 e^1 + ... + p_n e^n \mapsto p_j.$
The vector field $X_A$, which in these coordinates takes the form $$X_A(p) = A_{ij} x^j(p) \frac{\partial}{\partial x^i}_p$$ corresponds simply to the ODE $$\frac{d}{dt} x = Ax.$$
Now I want to understand what happens to the vector field $X_A$ under a change of coordinates of the form $$y = Cx.$$ In my intuition, this should correspond to a change of basis in the sense of linear algebra, that is I would expect in these new coordinates the vector field is of the form $X_B$, with $B = CAC^{-1}$.
However, when I do the calculations, I obtain a different results:
We have $$\frac{\partial}{\partial x^i} = \frac{\partial x^i}{\partial y^k}\frac{\partial}{\partial y^k},$$ hence since $x=C^{-1}y$, $$\frac{\partial}{\partial x^i} = C^{-1}_{ik} \frac{\partial}{\partial y^k}.$$
Plugging this into the definition of $X_A$, one obtains
$$X_A(p) = A_{ij} C^{-1}_{jl} y^l C^{-1}_{ik} \frac{\partial}{\partial y^k}= ({C^{-1}}^T AC^{-1})_{kl} y^l \frac{\partial}{\partial y^k} = X_{C^{-T}AC^{-1}}$$
This seems to suggest that the transformation behaviour is not as I thought above.
Now, am I making a mistake in the calculations here, or does a linear ODE under coordinate change not behave in the same way as a change of basis? If the latter, can someone give an intuitive explanation of the difference? Also, is there some transformation of ODE's that does behave in the way I expected?
Edit:
After reading the comments, I realize I might have been implicitly using some wrong assumption of independence of coordinates and basis, when simply replacing $x$ by $C^{-1}$ in the equation for the vector field instead of trying to solve for the coordinates in the new basis.
Thus, we posit $$X_A = A_{ij} x^j \frac{\partial}{\partial x^i} = d^k \frac{\partial}{\partial y^k}=d^k \frac{\partial y^k}{\partial x^i} \frac{\partial}{\partial x^i}=d^k C_{ki} {\partial x^i},$$ or, in matrix-vector notation
$$Ax \frac{\partial}{\partial x} = C^T d \frac{\partial}{\partial x}.$$
This should lead to $d = C^{-T}Ax$. But now we should be able to plug in $x = C^{-1}y$ here. Then the equation still reads
$$X_A = C^{-T}AC^{-1}y \frac{\partial}{\partial y}.$$