elliptic curves - Quadratic Twist of Legendre Form

To amplify on what Kevin Buzzard said: the Legendre form of an elliptic curve is

$E_{lambda} : y^2 = x(x-1)(x-lambda)$

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

$E_{lambda,d}: d y^2 = x(x-1)(x-lambda)$ rewritten into Legendre form. However, we do have

$j = 256 frac{(lambda^2 - lambda + 1)^3}{lambda^2 (lambda-1)^2}$

and there are six "versions" of $lambda$ which yield isomorphic curves in Legendre form:

$lambda,1-lambda,1/lambda,1/(1-lambda),lambda/(lambda-1),(lambda-1)/lambda$.

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 $Y^2 = X(X-a)(X-b)$ 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 2-torsion subgroup (satisfying a normalization of the Weil Pairing). In that case $a$ and $b$ are modular form of weight 2 for $Gamma(2)$