Consider a $n$-sided regular (convex) polygon and its circumscribed circle of radius $r$, centered in $(0,0)$. Fixing $(r,0)$ as the coordinate of the first vertex, the $n$ vertices of the polygon are given by:
$$P_i = (x_i,y_i)=(r\cos(2i\pi/n),r\sin(2i\pi/n))$$
The second moment of inertia w.r.t. the $x$-axis can then be derived using
$$ {\displaystyle I_{x}={\frac {1}{12}}\sum _{i=1}^{n}(y_{i}^{2}+y_{i}y_{i+1}+y_{i+1}^{2})(x_{i}y_{i+1}-x_{i+1}y_{i})}$$ leading to the result $I_x=nr^4/48\ (4\sin(2\pi/n)+\sin(4\pi/n))$.
Very interestingly, this value does not depend on the axis, or equivalently, on the rotation of the polygon: for example the two following hexagons have the same second moment of area wrt to the $x$-axis:
To show this, I multiplied $(x_i,y_i)$ by a rotation of matrix of angle $\theta$, leading to
$$P_i=(r\cos(2i\pi/n +\theta),r\sin(2i\pi/n + \theta)$$
Then, I computed $I_x(\theta)$ and using symbolic computation, observed that $I_x'(\theta)=0$.
Question Can the fact that $I_x$ does not depend on $\theta$ be seen easily by hand? Are there intuitive reasons for that?
Side note on mechanics This is interesting because it means the deflexion of a beam---the displacement induced by the force $F$, (see below)---with a regular-polygonial section does not depend on how the beam is placed. This is completely false for a rectangular section: the deflexion is much higher if the height of the beam corresponds the short side of the rectangle, as you can observe by holding a sheet of paper (slender rectangular section!) in a horizontal plane (huge deflection) or a vertical plane (no observable displacement).
The invariance is due to the regularity of the polygon, or rather due to the fact that there are more than two symmetry axes.
For an arbitrarily rotated cartesian coordinate system $r=(r_1,r_2)\in A$ of the cross-section, the second moments of area are components of a symmetric degree-$2$ tensor $$A_{ij} = \int_{r\in A} r_i r_j\,\mathrm{d}^2r\tag{1}$$ The area inertia tensor depends on $((A_{ij}))$ via $$J_{ij} = (A_{11}+A_{22}) \delta_{ij} - A_{ij} \tag{2}$$ where $\delta_{ij}$ is the Kronecker delta.
If one coordinate axis is a symmetry axis, then $(1)$ implies $A_{12} = 0$, that is, $((A_{ij}))$ is diagonalized, and the diagonal entries are the principal second moments of area. Thus a symmetry axis is a principal axis.
So, if there are more than two symmetry axes, there are more than two principal axes. That implies that $((A_{ij}))$ has an eigenspace of dimension more than one, which for a mere 2D cross-section implies $$A_{ij} = J_0\delta_{ij} \quad\text{for some scalar $J_0$}$$ and therefore $((A_{ij}))$ is invariant under rotation. Via $(2)$, this carries over to the area inertia tensor.