No. Given a hypergraph $H$, the $2$-shadow of $H$ is the $2$-uniform graph where we take each pair which is contained in an edge of $H$. This is isomorphism invariant.

Now consider the following graph on $\mathbb{Q}^2\cup (\mathbb{Q}\cup\{\infty\})$, where we think of $\mathbb{Q}\cup\infty$ as a slope. For each line in $\mathbb{Q}^2$, we draw an edge containing the points of the line and its slope; this is $H_1$. This is a complete regular linear hypergraph: obviously there are infinitely many lines through any given point, or of any given slope, and any two lines either have the same slope and no points in common, or otherwise different slopes and one point in common. The 2-shadow of this is a complete graph with all the edges between different slopes removed. In other words, it has one maximal independent set and every other pair is an edge.

Now we draw another graph $H_2$, whose vertices will be all rational lines in $\mathbb{Q}^3$ together with all rational normals in $\mathbb{Q}^3$. For each plane in $\mathbb{Q}^3$, we put an edge containing all the lines in that plane, together with the normal to the plane. Again, this is a complete regular linear hypergraph; for any line or normal, there are infinitely many planes through the line or with the normal, but any two planes either have the same normal and no line in common, or different normals and exactly one line in common.

But the 2-shadow of $H_2$ is more complicated than that of $H_1$. The set of normals is still an independent set, and it is a maximal independent set, because any line is contained in a plane which has a normal. But there are pairs of lines which are not in any plane (skew ones), so there are non-edges outside this maximal independent set.