Let $$E_1 : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$, $$a > b$$. Let $$E_2$$ be another ellipse such that it touches the end points of major axis of $$E_1$$ and the foci of $$E_2$$ are the end points of minor axis of $$E_1$$. If $$E_1$$ and $$E_2$$ have same eccentricities, then its value is:

For ellipse $$E_1: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ with $$a > b$$, the major axis endpoints are $$(\pm a, 0)$$, the minor axis endpoints are $$(0, \pm b)$$, and the eccentricity is $$e$$ where $$b^2 = a^2(1 - e^2)$$ $$-(1)$$.

Ellipse $$E_2$$ touches the endpoints of the major axis of $$E_1$$, so $$E_2$$ passes through $$(\pm a, 0)$$. The foci of $$E_2$$ are the endpoints of the minor axis of $$E_1$$, i.e., $$(0, \pm b)$$.

Since the foci of $$E_2$$ lie on the $$y$$-axis, the major axis of $$E_2$$ is along the $$y$$-axis. Let $$E_2: \frac{x^2}{B^2} + \frac{y^2}{A^2} = 1$$ where $$A > B$$.

The foci of $$E_2$$ are at $$(0, \pm c_2)$$ where $$c_2^2 = A^2 - B^2$$. Since the foci are at $$(0, \pm b)$$, we get $$A^2 - B^2 = b^2$$ $$-(2)$$.

Since $$E_2$$ passes through $$(\pm a, 0)$$: $$\frac{a^2}{B^2} + 0 = 1$$, so $$B = a$$ $$-(3)$$.

From $$(2)$$ and $$(3)$$: $$A^2 = a^2 + b^2$$ $$-(4)$$.

The eccentricity of $$E_2$$ is $$e_2 = \frac{b}{A} = \frac{b}{\sqrt{a^2 + b^2}}$$.

Setting $$e_1 = e_2$$: $$\frac{\sqrt{a^2 - b^2}}{a} = \frac{b}{\sqrt{a^2 + b^2}}$$.

Squaring: $$\frac{a^2 - b^2}{a^2} = \frac{b^2}{a^2 + b^2}$$.

Cross-multiplying: $$(a^2 - b^2)(a^2 + b^2) = a^2 b^2$$, so $$a^4 - b^4 = a^2 b^2$$.

Dividing by $$a^4$$: let $$t = \frac{b^2}{a^2}$$, then $$1 - t^2 = t$$, so $$t^2 + t - 1 = 0$$.

Solving: $$t = \frac{-1 + \sqrt{5}}{2}$$ (taking the positive root).

Now $$e^2 = 1 - t = 1 - \frac{-1 + \sqrt{5}}{2} = \frac{3 - \sqrt{5}}{2}$$.

We can verify that $$e = \frac{-1+\sqrt{5}}{2}$$ satisfies $$e^2 = \frac{3 - \sqrt{5}}{2}$$: indeed $$\left(\frac{-1+\sqrt{5}}{2}\right)^2 = \frac{1 - 2\sqrt{5} + 5}{4} = \frac{6 - 2\sqrt{5}}{4} = \frac{3 - \sqrt{5}}{2}$$. This confirms $$e = \frac{-1+\sqrt{5}}{2}$$.

The answer is $$\frac{-1+\sqrt{5}}{2}$$, which is Option A.