Let $a$ be the hypotenuse and $b,c$ the others sides, then by Pythagorean Theorem
$$a^2=b^2+c^2.$$
Then
$$a^2>b^2,\quad a^2>c^2.$$
Therefore
$$a>b\quad a>c.$$
I think I have a solution but it will be difficult to explain. I will use the following variables:
$$
\begin{split} a&=\text{base of orange triangle} \\ b&=\text{height of orange triangle} \\ c&= \text{"left" side of green triangle} \end{split}
$$
Then the hypotenuse of the orange triangle is $\sqrt{a^2+b^2}$ and the hypotenuse of the green triangle is $\sqrt{a^2+b^2+c^2}$. Since the perimeters of both triangles are equal we must have:
$$
\begin{split} &&a+b=c+\sqrt{a^2+b^2+c^2} \\ &\implies& a+b-c=\sqrt{a^2+b^2+c^2} \\ &\implies& (a+b-c)^2=a^2+b^2+c^2 \\ &\implies& a^2+b^2+c^2+2ab-2ac-2bc=a^2+b^2+c^2 \\ &\implies& ab=ac+bc \\ &\implies& c=\frac{ab}{a+b} \end{split}
$$
So now we have all sides in terms of $a$ and $b$. But we can relate $a$ and $b$ using the angle we are given. So:
$$
\tan(20)=\frac{b}{a} \implies b=a\tan(20)
$$
Using this we get:
$$
\begin{split} a^2+b^2=a^2+a^2\tan^2(20)=a^2(1+\tan^2(20))=\frac{a^2}{\sin^2(20)} \end{split}
$$
Now we can get an expression for $a^2+b^2+c^2$ purely in terms of $a$:
$$
\begin{split} a^2+b^2+c^2&= \frac{a^2}{\sin^2(20)}+\frac{a^2b^2}{\frac{a^2}{\sin^2(20)}+2ab}\\ &=\frac{a^2}{\sin^2(20)}+\frac{a^4\tan^2(20)}{\frac{a^2}{\sin^2(20)}+2a^2\tan(20)} \\ &=\frac{a^2}{\sin^2(20)}+\frac{a^4\tan^2(20)\sin^2(20)}{a^2+2a^2\tan(20)\sin^2(20)} \\ &=\frac{a^2}{\sin^2(20)}+\frac{a^2\tan^2(20)\sin^2(20)}{1+2\tan(20)\sin^2(20)} \\ &=a^2\left[ \frac{1}{\sin^2(20)}+\frac{\tan^2(20)\sin^2(20)}{1+2\tan(20)\sin^2(20)} \right] \\ &= a^2K^2 \end{split}
$$
where $K^2$ is just the number in the square brackets (the only reason I made it $K^2$ instead of $K$ is because we will need to take square roots). Finally, letting $\theta$ be the smallest angle of the green triangle, we get:
$$
\begin{split} \cos(\theta)&=\frac{\sqrt{a^2+b^2}}{\sqrt{a^2+b^2+c^2}} \\ &=\frac{\frac{a}{\sin(20)}}{aK} \\ &=\frac{1}{K\sin(20)} \end{split}
$$
You can work out that $K\approx 2.9262$ and then plugging this in gives us $\theta\approx 2.34^{\circ}$.
Best Answer
I'll work through a slightly-more-general problem, replacing $20^\circ$ with $\alpha$. Writing $d$ for the hypotenuse of the lower triangle, and $\beta$ for the angle shown in the upper triangle ...
... the equal perimeters condition tells us:
$$d + d \sin\alpha + d \cos\alpha = d + d\tan\beta + d\sec\beta \tag{1}$$
That is,
$$\sin\alpha + \cos\alpha - \tan\beta = \sec\beta \tag{2}$$
Squaring both sides of (2), and noting the $\sec^2\theta = \tan^2\theta + 1$, we have $$( \sin\alpha + \cos\alpha )^2 - 2 \tan\beta ( \sin\alpha + \cos\alpha ) + \tan^2\beta = \tan^2\beta + 1 \tag{3}$$ so that $$\begin{align} 2\tan\beta (\sin\alpha + \cos\alpha) &= ( \sin\alpha + \cos\alpha )^2 - 1 \tag{4}\\ &= \sin^2\alpha + \cos^2\alpha + 2 \sin\alpha \cos\alpha - 1 \tag{5}\\ &= 1 + 2\sin\alpha\cos\alpha - 1 \tag{6}\\ &= 2\sin\alpha\cos\alpha \tag{7} \end{align}$$
Therefore,
In the particular case that $\alpha = 20^\circ$, we have $$\tan\beta = 0.250753\ldots \quad\to\quad \beta = \operatorname{atan} 0.25073\ldots = 14.0769\ldots^\circ$$
I don't believe there's a "nice" form for this angle.