By switching integrals in double integral, I showed that $$\int_0^1K(k)dk=\int_0^1K'(k)dk=2G$$ where $K(k)$ is complete elliptic integral of the first kind and $K'(k)=K(\sqrt{1-k^2})$ is its complement.
But, I couldn't show the equality $$\int_0^1K^2(k)dk=\frac12\int_0^1K'^2(k)dk$$ and no idea about its value.
I am extremely weak in these topics. Thanks for your kind comments and answers.
Set $k=\frac{1-u}{1+u}$ on the left-hand side and apply the following quadratic transform.
$$K'(k)=\frac{2}{1+k}K\left(\frac{1-k}{1+k}\right)\tag{1}$$
We have,
$k+uk=1-u\implies u=\frac{1-k}{1+k}$
And, $\mathrm dk=-\frac{2}{(1+u)^2}\mathrm du$.
Now, $$\int_0^1K^2(k)~\mathrm dk=-\frac12\int_1^0K^2\left(\frac{1-u}{1+u}\right)~\frac{4\mathrm du}{(1+u)^2}=\frac{1}{2}\int_0^1K'^2(k)~\mathrm dk\qquad\text{proved.}$$
In general, applying the above quadratic transform along with $K(k)=\frac{1}{1+k}K\left(\frac{2\sqrt k}{1+k}\right)$ leads to $$\int_0^1K^n(k)~\mathrm dk=\frac12\int_0^1K'^n(k)\left(\frac{1+k}2\right)^{n-2}~\mathrm dk$$
$$\int_0^1K'^n(k)~\mathrm dk=2\int_0^1K^n(k)\left(1+k\right)^{n-2}~\mathrm dk$$
I assume you're using the definition $$K(k) = \int_{0}^{\pi/2} \frac{\mathrm d \theta}{\sqrt{1-k^{2} \sin (\theta)}}. $$
Contour integration shows that $$\int_{0}^{1} K(k)^{2} \, \mathrm dk + \int_{1}^{\infty} \left(\frac{1}{k} \, K \left(\frac{1}{k} \right)+ i K(\sqrt{1-k^{2}}) \right)^{2} \, \mathrm dk - i \int_{0}^{\infty}K(ik)^2 \, \mathrm dk = 0.$$
This equation is obtained by integrating the principal branch of $K(k)^2$ around the indented quarter-circle contour $$[0,1-\epsilon] \cup \epsilon e^{i[\pi, o]} \cup [1+ \epsilon, R] \cup Re^{i [0, \pi/4}] \cup [iR, 0],$$ where the integration on $[1+ \epsilon, R]$ is on the upper side of the branch cut.
See this related answer for more details. Notice that I'm using the definition $$K(m) = \int_{0}^{\pi/2} \frac{\mathrm d \theta}{\sqrt{1-m \sin^{2}(\theta)}}$$ in that other answer. That's the definition that Mathematica/WolframAlpha uses.
Equating the real parts on both sides of the equation, we have $$ \begin{align} 0 &= \int_{0}^{1} K(k)^{2} \, \mathrm dk + \int_{1}^{\infty} \frac{1}{k^{2}} \, K \left(\frac{1}{k} \right)^{2} \, \mathrm dk - \int_{1}^{\infty} K \left(\sqrt{1-k^{2}} \right)^{2} \, \mathrm dk \\ &= \int_{0}^{1} K(k)^{2} \, \mathrm dk + \int_{0}^{1} K(u)^{2} \, \mathrm du - \int_{0}^{1} \left( \frac{1}{u} \, K \left(\sqrt{1-\frac{1}{u^{2}}}\right) \right)^{2} \, \mathrm du \\ & \overset{\clubsuit}{=} 2 \int_{0}^{1} K(k)^{2} \, \mathrm dk - \int_{0}^{1} K \left(\sqrt{1-u^{2}} \right)^{2} \, \mathrm du. \end{align} $$
Therefore, $$\int_{0}^{1} K(k)^{2} \mathrm dk = \frac{1}{2} \int_{0}^{1} K \left(\sqrt{1-u^{2}} \right)^{2} \, \mathrm du. $$
$\clubsuit$ See $\spadesuit$ in my previous answer.
Integrating $K(k)^3$ around the same contour shows that $$\int_{0}^{1} (k+1) K(k)^{3} \, \mathrm dk = 3 \int_{0}^{1} k K(k) K \left(\sqrt{1-k^{2}} \right)^{2} \, \mathrm dk. $$
