Let $E/k$ be an elliptic curve defined by the Weierstrass form $y^2=x^3+ax+b$. Let $c$ be a nonzero square free element in $k$. Let ${E_c}/k$ be a curve defined by $cy^2=x^3+ax+b$. Using a linear change of variables in $\bar{k}$ (isomorphism over $\bar{k}$), show that $E_c$ is a twist of $E$.
So I want to show that $E_c$ and $E$ are isomorphic over $\bar{k}$ but not isomorphic over $k$. I think the best way to do this is to show they have the same j-invariant. However, I'm not sure what change of variables to use to show that.
What's the best way to show $E_c$ is a twist of $E$?
Per your request, I am turning my comment into an answer:
Consider the change of variables \begin{align*} x' & = x\\ y' &= \sqrt{c}y \end{align*} Then in the new variables we have $E_c: (y')^2 = (x')^3 + ax' + b$. The map $y\mapsto y'$ is multiplication by a non-zero element of $\overline{k}$ which gives a $\overline{k}$ isomorphism between $E$ and $E_c$.
Note: This map is defined over $k(\sqrt{c})$; thus the two curves actually become isomorphic over a quadratic extension.