Let $f: \mathbb{R}^n \rightarrow \mathbb{R}$ where $y=f(x)$ and $x=[x_1,\dots, x_n]^T$. Suppose we reparametrize $f(x)$ such that $z_i=c_ix_i$ to define a new scalar-valued function $g=f(z)$ where $z=[z_1,\dots, z_n]^T=[c_1x_1,\dots, c_nx_n]^T$.
Question: How can one write $\nabla_c(f(z))\mid_{c=[1,\dots,1]^T}$ based on the original function $f(x)$?
My try:
One can write the jacobian of $f(z)$ as follows:
$$ \nabla^T_c(f(z))=J_cf(z)=J_zf(z)J_c(z)=J_zf(z) \begin{bmatrix} x_1&0&\cdots&0\\ 0 & x_2 & & \vdots\\ \vdots & & \ddots&0\\ 0 & \cdots & 0 &x_n \end{bmatrix}=J_zf(z)\text{diag}(x) $$ Hence, $\nabla_c(f(z))=\text{diag}(x)J^T_zf(z)$.
Here is where I am not sure: $\nabla_c(f(z))\mid_{c=[1,\dots,1]^T}=\text{diag}(x)J^T_zf(z)\mid_{c=[1,\dots,1]^T}=\text{diag}(x)J^T_xf(x)$. How one rigorously can show they are equal?
I suspect that $J^T_zf(z)\mid_{c=[1,\dots,1]^T}=J^T_xf(x)$ because considering the scalar $x$ and $c$ along with function $g=\sin(cx)$ one has $\frac{\partial \sin(cx)}{\partial c}|_{c=1}=x\cos(x)$.