The rescalling $(V')$ of $(V)$ as described in the question has always an isochronous center when $(V)$ is a quadratic system with center.

The reason is that $d\theta(V')=1$ where $d\theta=(\frac{1}{x^2+y^2})(ydx-xdy)$.

The rescalling $(V')$ of $(V)$ as described in the question has always an isochronous center when $(V)$ is a quadratic system with center.

The reason is that $d\theta(V')=1$ where $d\theta=(\frac{1}{x^2+y^2})(ydx-xdy)$.

Remark: The same argument as above can be applied to show the following:

Let $V$ be an arbitrary quadratic vector field on the plane and $(V')$ be the corresponding rescalling as in the question. (We no longer assume that $V$ has necessarily a center). Then all closed orbits or limit cycles of $V'$ which surround the origin have the same length provided we choose a Riemannian metric whose frame is in the form $\{V', f(x\partial_x+y\partial_y)\}$ where $f$ is an arbitrary positive smooth function.

The rescalling of $(V)$ as described on the question has always an isochronous center when $(V)$ is a quadratic with center.

The reason is that if we denote this rescalling by $(V')$, then $d\theta(V')=1$ where $d\theta=(\frac{1}{x^2+y^2})(ydx-xdy)$.

The rescalling $(V')$ of $(V)$ as described in the question has always an isochronous center when $(V)$ is a quadratic system with center.

The reason is that   $d\theta(V')=1$ where $d\theta=(\frac{1}{x^2+y^2})(ydx-xdy)$.