Let $J = J(\tau)$ be Klein's invariant and let $0 < k < 1$ be the elliptic modulus. It is known that $$J = \frac{4}{27} \frac{(1 - \lambda + \lambda^2)^3}{\lambda^2 (1 - \lambda)^2},$$ where $\lambda = k^2$. However, when the discriminant $\Delta$ is negative, $J(\tau)$ changes into $$J' = -\frac{(1 - 16\lambda + 16\lambda^2)^3}{108\lambda(1 - \lambda)}.$$ Furthermore, it can be shown that $$J(2\tau) = \frac{(16 - 16\lambda + \lambda^2)^3}{108\lambda^4 (1 - \lambda)}.$$ Incidentally, the right-hand side is the normalized $j$-invariant of the elliptic curve $$E_1 : y^2 = (1 - t^2)(t^2 - \lambda'), \quad \lambda' = (1 - \lambda).$$ Moreover, we have $$J\bigg(\frac{\tau}{2}\bigg) = \frac{(1 + 14\lambda + \lambda^2)^3}{108\lambda (1 - \lambda)^4},$$ where the right-hand side is the normalized $j$-invariant of $$E_2 : y^2 = (1 - t^2)(1 - \lambda t^2).$$ On the other hand, $J'$ also happens to be the normalized $j$-invariant of the elliptic curve $$E_3 : y^2 = (1 - t^2)(\lambda' + \lambda t^2).$$ This raises the question: Does $J'$ equal $J(r(\tau))$, where $r(\tau)$ is some simple rational function of $\tau$?
Edit: It appears that $$J' = J\bigg(\frac{\tau + 1}{2}\bigg).$$ I have a proof, but this margin is too small to contain it.
