This is not a complete answer, but a mathematical analysis of the problem.
Since we can scale the graph, we may look for solutions in rational numbers rather than in integers.
Without loss of generality, we may assume that one triangle has side length $1$. If we let $a$ be the side length of the other triangle and let $\theta$ be the angle of rotation between them, then we are facing the following conditions:
\begin{eqnarray} \frac 1 3 \left(1 + a^2 - 2a\cos \theta\right) &\in& \Bbb Q^2\\ \frac 1 3 \left(1 + a^2 - 2a\cos (\theta + \frac {2\pi} 3)\right) &\in& \Bbb Q^2\\ \frac 1 3 \left(1 + a^2 - 2a\cos (\theta - \frac {2\pi} 3)\right) &\in& \Bbb Q^2\\ \end{eqnarray}
These conditions implie that $\cos \theta$ and $\cos(\theta \pm \frac {2\pi}3)$ are all rational numbers.
It is easy to parametrize the possible values of $\cos \theta$ fulfilling the above condition.
If we write $c$ for $\cos \theta$ and $s$ for $\sin \theta$, then the conditions are: \begin{eqnarray} c &\in& \Bbb Q\\ -\frac 1 2 c \pm \frac {\sqrt 3}2 s &\in& \Bbb Q\\ c^2 + s^2 &=& 1 \end{eqnarray} Writing $d = s /\sqrt 3$, we are simply looking for $c, d \in \Bbb Q$ such that $c^2 + 3d^2 = 1$.
A standard procedure (projecting from the point $(c, d) = (1, 0)$) then gives us $c = \frac {t^2 - 3}{t^2 + 3}$ for some $t \in \Bbb Q$.
Therefore, the original conditions translate to the following diophantine equation: \begin{eqnarray}\tag{*} 1 + a^2 - a\cdot \frac{2t^2 - 6}{t^2 + 3} &=& 3u^2\\ 1 + a^2 - a\cdot \frac{-t^2 + 6t + 3}{t^2 + 3} &=& 3v^2\\ 1 + a^2 - a\cdot \frac{-t^2 - 6t + 3}{t^2 + 3} &=& 3w^2\\ \end{eqnarray} We want to find rational solutions $(a, t, u, v, w)$ (with some conditions) to this group of equations.
This is, of course, an algebraic surface. I have almost no knowledge about rational points on a general algebraic surface, especially such a complicated one.
I would be very much impressed if the OP has a complete answer to this question. And I think the question is probablly more suitable for the MathOverflow site.
Below I give some observations.
If we put $a = 1$, then we cut out a curve on the surface. This curve is very special:
\begin{eqnarray} \frac{4}{t^2 + 3} &=& u^2\\ \frac{(t - 1)^2}{t^2 + 3} &=& v^2\\ \frac{(t + 1)^2}{t^2 + 3} &=& w^2\\ \end{eqnarray} The mapping $(t, u, v, w)\mapsto (t, 2/u)$ then gives (up to irreducible components) a birational equivalence of this curve with the curve $x^2 + 3 = y^2$, which is again birationally equivalent to the projective line.
This implies that there are infinitely many rational points on this curve, which can be given by explicit formulas with one rational parameter.
However, these are the solutions that the OP excludes.
For any fixed (positive) $a\neq 1$, each single equation in $(*)$ becomes a curve of genus $1$, i.e. it is (geometrically) an elliptic curve. This suggests that rational points might be very rare, as we are looking for a value of $t$ that simultaneously gives three rational points on three elliptic curves.
However, this is just a heuristic argument and might not be true. For example, if we replace $3u^2, 3v^2, 3w^2$ with $u^2, v^2, w^2$, then there are indeed many more rational points away from $a = 1$. My guess is that they come from another rational curve on the surface.
The final conclusion of this answer is that this problem is too difficult for me, as I don't have enough knowledge to solve it. I also have no idea what are the known results on this subject (if this is posted on the MathOverflow site, then we can expect experts telling us the state of art).
Of course, the OP only asks for one non-trivial rational point (which in some sense becomes a programming task). But as stated above, I would be very impressed if one can actually determine all the rational points, or obtain any non-trivial interesting results.