Is it possible to inscribe a regular hexagonal pyramid inside a sphere, so that all corners are touching the pyramid? Since the height of the pyramid is'nt necessarily the same as the lenght of the edge, I can't figure out if it is even possible.
Here is a picture of the solution I want, but made with a tetragonal pyramid:
On
The base of a regular hexagonal pyramid can be inscribed on a circle, with radius of the edge length, whose plane has a normal through the circle's centre and the apex.
Then the pyramid can be rotated around that normal and all basal vertices will move along the circle. Thus the problem is equivalent to finding a cone with its apex and base boundary both on the same sphere.
But then this is merely the surface resulting from an isosceles triangle being rotated around a line bisecting its base, where its three vertices are all on a circle. Since three points define a circle, and this triangle-circle combination can be scaled as necessary, you can always find such a circle for any given triangle or vice versa.
Therefore you can find the cone and thus the hexagonal pyramid, or the corresponding sphere to fit around them.
Yes, it is possible. I will give you the general statement, so you would know when any pyramid, not just a regular one, is inscribed in a sphere.
Statement. Let $AB_1B_2...B_n$ be a pyramid (not necessarily regular) with apex $A$ and base planar polygon $B_1B_2...B_n$, where in your case $n=6$. The pyramid $AB_1B_2...B_n$ can be inscribed in a sphere if and only if the base polygon $B_1B_2...B_n$ can be inscribed in a circle.
Proof. Assume that $B_1B_2...B_n$ can be inscribed in a circle, i.e. there exists a circle with center $O'$ such that all vertices $B_1, B_2, ..., B_n$ lie on the circle. Let $l$ be the line through the center point $O'$ and orthogonal to the plane determined by the base polygon $B_1B_2...B_n$.
Denote by $M$ the midpoint of the edge $AB_1$ and let $p$ be the plane through $M$ and orthogonal to $AB_1$, i.e. this is the plane of all orthogonal bisector lines of segment $AB_1$.
Denote by $O$ the unique intersection point of the plane $p$ with the line $l$. Then $OA = OB_1$ because triangles $AMO$ and $B_1MO$ are congruent. Indeed, $MO$ is a common segment, $AM = B_1M$ because $M$ is a midpoint of $AB_1$ and $\angle \, AMO = \angle \, B_1MO = 90^{\circ}$ because plane $p$, containing $MO$, is orthogonal to $AB_1$.
Furthermore, for $O$ on $l$ the distances $OB_1=OB_2= ... =OB_n$ because any pair of triangles $OO'B_1$ and $OO'B_k$ are congruent for all $k=2,...,n$. Indeed, $OO'$ is a common edge, $O'B_1 = O'B_k$ as radii of the circle circumscribed around $B_1B_2...B_n$ with center $O'$, and $\angle \, OO'B_1 = \angle \, OO'B_k = 90^{\circ}$ because $OO' = l$ is orthogonal to the plane of the base polygon.
Consequently, $OA = OB_1 = OB_2 = ... = OB_n $ which means that the points $A, B_1, B_2, ... , B_n$ are equidistant from the common point $O$, which means they all lie on the sphere centered at $O$ and of radius $OA$.
The converse statement is trivial, if the pyramid $AB_1B_2...B_n$ is inscribed in a sphere, the planar polygon $B_1B_2...B_n$ determines a plane which intersects the sphere in a circle and all of the vertices of $B_1B_2...B_n$ lie simultaneously on the plane and on the sphere, so they lie on a common circle.