Icosahedron if rotated

$\Large \displaystyle \text{Solution }1$ Euler's Formula tells us that $V - E + F = 2$ , where an icosahedron has $V$ vertices, $E$ edges, and $F$ faces. We're told that $F = 20$ . Each triangle has 3 edges and every edge is common to 2 triangles, so $E = \dfrac{20 \times 3}{2} = 30.$ Additionally, each triangle has 3 vertices, so if every vertex is common to n triangles, then $V = \dfrac{20 \times 3}{n} = \dfrac{60}{n}$ . Plugging this into the formula, we have

$\dfrac{60}{n} - 30 + 20 = 2 \implies \dfrac{60}{n} = 12 \implies n = 5$

Again this shows that the rotation is $\dfrac{360^{\circ}}{5} = \color{#D61F06}{\boxed{72^{\circ}}}$ .

$\Large \displaystyle \text{Solution }2$

Because this polyhedron is regular, all vertices must look the same. Let’s consider just one vertex. Each triangle has a vertex angle of $60^{\circ}$ , so we must have fewer than $6$ triangles; if we had $6$ , there would be $360^{\circ}$ at each vertex and you wouldn’t be able to fold the polyhedron up. It’s easy to see that we need at least 3 triangles at each vertex, and this gives a triangular pyramid with only 4 faces. Having 4 triangles meeting at each vertex gives an octahedron with 8 faces. Therefore, an icosahedron has 5 triangles meeting at each vertex, so rotating by $\dfrac{360^{\circ}}{5} = \color{#D61F06}{\boxed{72^{\circ}}}$ gives another identical icosahedron.

Five triangles meet at the vertex. So rotating by 360/5= $\Large \color{#D61F06}{72^o}$ , each triangle would fit exactly into its next forward friend.