For a function $f\in C^2$, $f:\mathbb{R}^n\to\mathbb{R}$ and a point $x\in\mathbb{R}^n$ with $\nabla^2f(x)$ positive definit one can calculate the new point $x^+=x+s$ as follows:
Change the coordinate system: $x=x(y)=My+v$, $y\in\mathbb{R}^n$; $M\in\mathbb{R}^{n\times n}$ invertible and $v\in\mathbb{R}^n$.
Gradient step in the $y$-coordinate-system: $y^+=y-\nabla_yf(x(y))$.
Retransformation into $x$-coordinates: $x^+=x(y^+)$.
The task is to find a formula for the step $s$ which only depends on $M$ and $\nabla f(x)$.
I tried the following:
$s=x^+-x=x(y^+)-x(y)=My^++v-My-v=M(y^+-y)=M(y-\nabla_yf(x(y))-y)=-M\nabla_yf(x(y))$
This formula depends on $\nabla_yf(x(y))$. The question is how $\nabla_y$ can be transformed to $\nabla$.