What is the quadratic twist of an elliptic curve in Legendre Form? How do you show an elliptic curve and its quadratic twist is isomorphic when they are in Legendre Form?

To amplify on what Kevin Buzzard said: the Legendre form of an elliptic curve is $E_{\lambda} : y^2 = x(x1)(x\lambda)$ Since it has only one parameter, it has much less freedom than $y^2 = x^3 + ax + b$. Strictly speaking, an elliptic curve over a field $K$ is a twist of another elliptic curve over $K$, if they are not isomorphic over $K$ but they are over an extension $L/K$. So it doesn't make sense to ask when the quadratic twist (meaning that $K=\mathbb{Q}$ and $L=\mathbb{Q}(\sqrt{d})$ is a quadratic extension) is isomorphic to the curve. However, I think that you mean the curve $E_{\lambda,d}: d y^2 = x(x1)(x\lambda)$ rewritten into Legendre form. However, we do have $j = 256 \frac{(\lambda^2  \lambda + 1)^3}{\lambda^2 (\lambda1)^2}$ and there are six "versions" of $\lambda$ which yield isomorphic curves in Legendre form: $\lambda,1\lambda,1/\lambda,1/(1\lambda),\lambda/(\lambda1),(\lambda1)/\lambda$. This seriously restricts the possible $d$'s for which $E_{\lambda,d}$ can be written in Legendre form to a at most six values, namely if $\pm d \lambda, \pm d, \pm d (1\lambda)$ are squares. [Added later]: It might make more sense to think of $Y^2 = X(Xa)(Xb)$ as a "generalized Legendre form". In that line of thought, the "standard" Weierstrass form $Y^2 = X^3 + a X + b$ is associated with the moduli space $X(1)$, and $a$ is a modular form for weight 4, and b is a modular form of weight 6 for $\Gamma(1)$ (suitably normalized Eisenstein forms). The "generalized Legendre form" is associated with $X(2)$ which parametrizes elliptic curves along with the full 2torsion subgroup (satisfying a normalization of the Weil Pairing). In that case $a$ and $b$ are modular form of weight 2 for $\Gamma(2)$ 

