General means of obtention of regular tetrahedra inscribed into a regular tetrahedron This is a follow-up of a recent question that I will name $(Q)$.
Reformulating question $(Q)$: being given a regular tetrahedron $(T)=(ABCD)$, how is it possible to inscribe (slanted) smaller regular tetrahedra $A'B'C'D'$  inside $(T)$ (i.e., with $A' \in BCD, B' \in ACD...$.
An elegant solution is given there (with alas few details) using an adequate rotation combined with a shrinking factor $1/3<f<1$.
This solution did not fully satisfied me because I want to know if there other means to get inscribed tetrahedra.
I have taken the vetices of the regular tetrahedron $(T)$ as the columns of this matrix:
$$T=\left(\begin{array}{rrrr}1 &-1& 1& -1\\1& 1 &-1 &-1\\1 &-1 &-1 &1\end{array}\right)\tag{1}$$
(obtained by taking a system of 4 non-neighbor vertices of a cube).
Here are the equations of the planes containing the faces of tetrahedron $(T)$:
$$\begin{cases}x+y+z+1&=&0\\-x+y+z-1&=&0\\x-y+z-1&=&0\\x+y-z-1&=&0\end{cases}\tag{2}$$
I decided to use a CAS (Computer Algebra System).
After a certain number of failures, I have finaly obtained a satisfying way to get particular solutions. Here is the central part of the corresponding (Matlab CAS) program on which I am going to make some comments:
syms x1 x2 x3 x4 y1 y2 y3 y4 z1 z2 z3 z4
d12=(x1-x2)^2+(y1-y2)^2+(z1-z2)^2;
d13=(x1-x3)^2+(y1-y3)^2+(z1-z3)^2;
d14=(x1-x4)^2+(y1-y4)^2+(z1-z4)^2;
d23=(x2-x3)^2+(y2-y3)^2+(z2-z3)^2;
d24=(x2-x4)^2+(y2-y4)^2+(z2-z4)^2;
d34=(x3-x4)^2+(y3-y4)^2+(z3-z4)^2;
    [X1,X2,X3,X4,Y1,Y2,Y3,Y4,Z1,Z2,Z3,Z4]=...
    solve(...
      x1+y1+z1+1==0,x2-y2+z2-1==0,x3-y3-z3+1==0,x4+y4-z4-1==0,...
      x1+x2+x3+x4==0,y1+y2+y3+y4==0,z1+z2+z3+z4==0,...
      d12==d13,d12==d14,d14==d23,d23==d24,...
      z1==-1/3,...
    x1,x2,x3,x4,y1,y2,y3,y4,z1,z2,z3,z4)

Let me explain the different lines following the "solve" instruction, being understood that the vertices of the looked for tetrahedra are $(x_k,y_k,z_k)$:

*

*first line: the vertices of tetrahedra  $(T')$ must belong to the faces (see (2)),


*second line: the center of $(T')$ must be the origin,


*third line: $(T')$ must be regular,


*fourth line: one of the coordinates of the first vertex of $(T')$ is fixed arbitrarily.
In the case at hand, we obtain almost instantly different solutions. Among them, in blue, we obtain the very particular "central" tetrahedron whose vertices are the centroids of the faces of $(T)$ (this is why we had taken the particular value $z_1=-1/3$...). The two other solutions (in red and green) are spurious, but fulfill all the conditions...

Fig. 1: Results of the program given above.

Fig. 2: With a different value of $z_1$, four different solutions for $T'$.

Fig. 3: The  vertices of many different inscribed tetrahedra $(T')$ for many different values of $z_1$, making apparent the fact that they belong to hyperbolic arcs as shown in the answer by @Intelligenci Pauca.
Remark: when a solution $(T')$ has been found (as before, we assimilate a tetrahedron with the matrix $3 \times 4$ of the coordinates of its vertices), we can retrieve

*

*the scaling factor $s$ by computing the ratio of norms of the columns of matrices $(T)$ and $(T')$, and


*the rotation $R$ such that $T'=sRT$ by computing:
$$R=\frac1s T'T^+$$
where $T^+$ is the $4 \times 3$ pseudo-inverse of $T$.

My question is twofold:

*

*With my method based on vertices of $(T')$, are we able to retrieve a larger set of combinations "rotation + shrinking" than in question $(Q)$ ? Are there general formulas that avoid to consider particular cases ?


*Same question as in $(Q)$, to which no answer has been given: is there some interesting papers about the issue of tetrahedra inscribed in a tetrahedron ?
Edit: This question has found a new direction with the answer by Intelligenci Pauca that I have attempted to explain in a simple way (see my answer).
 A: This is not an answer.  Just a visualization of the similar regular tetrahedron being rotated and remaining inscribed in the original tetrahedron, as the angle of rotation changes from $0$ to $\pi$.
To generate the view, I used a regular tetrahedron centered at the origin with one of its faces parallel to the $xy$ plane.  A copy of that tetrahedron is generated, and its frame (the frame of reference attached to it) is rotated about an axis that passes through the origin and extending between the midpoint of one edge and the midpoint of the opposing edge as first suggested by Jaap Scherphius in his comment to  my recent question.
The tetrahedron copy is then scaled uniformly by the scale factor
$ f = \dfrac{1}{2 |\sin \theta| + 1 } $
and this was also suggested by Jaap Scherphuis in his comment, where $\theta$ is the angle of rotation.
The curve in red is the trace of one of the vertices of the smaller tetrahedron as it rotates.  After the small tetrahedron completes its rotation, the tetrahedrons (at that moment they are coincident) are rotated so that the plane containing the red curve faces the viewer.
The .GIF animation was created using my own MS-Excel software that uses VBA (Visual Basic for Applications) script.  The generated Excel Chart frames were put together into a .GIF image online using this website

A: This is not an answer. I just want to show the locus of vertex $K$ (or $N$) of inscribed tetrahedron $KNQR$, obtained by the rotation of a a tetrahedron similar to $ABCD$, with the same centre $O$ and a vertex at $E$, about the line $r$ passing through the midpoints of $AB$ and $CD$.
The locus is the red arc, passing through $C$, $D$ and the centre $G$ of face $BCD$ (not shown). In blue, the parabola through $C$, $D$ with vertex in $G$. Of course the complete locus contains two other arcs, through $BGC$ and through $DGB$.

EDIT.
The locus is of course the intersection between a cone (having $OC$, $OD$ as generatrices) and plane $BCD$, i.e. a hyperbola. That cone is the locus of the circles generated by the rotation of $E$ about $r$, as $E$ varies on $OD$.
In figure below one can see a section of the cone, perpendicular to the intersecting plane $CDG$ ($G$ is the centre of face $BCD$), which is represented by dashed line $GL$. If we take $OC=OD=1$ then $TT'=CD={2\over3}\sqrt6$ and $OG=1/3$.
Points $G$ and $L$ are the vertices of the hyperbola: a little geometry shows that $OL=1$, so that the major axis is
$GL=2a={2\over3}\sqrt2=GB$. Hence $L\equiv B$. Moreover, $LL'=TT'={2\over3}\sqrt6$ and
$GG'={1\over3}TT'={2\over9}\sqrt6$, hence the secondary axis is given by (see here for details):
$$
2b=\sqrt{LL'\cdot GG'}={2\over3}\sqrt2=2a.
$$
This is then a rectangular hyperbola.

A: After some hesitations, instead of writing an ever inflating "Edit" to my question, here is a personal answer. Indeed, in the light of the answer by @Intelligenci Pauca, I have realized that this issue should be placed into a different, rather simpler, context.
Let us call $(T)$ the reference regular tetrahedron and $(T')$ its "middle tetrahedron", whose vertices are the centers of the faces of $(T)$.
In fact, the different tetrahedra that can be inscribed into $(T)$ are obtained as "steps" in the four continuous transformations (rotation $R_t$ + shrinking by factor $f$) bringing $(T)$ onto $(T')$, obtained for example with this vertical axis rotation:
$$\underbrace{\left(\begin{array}{rrrr}x_1&x_2&x_3&x_4\\y_1&y_2&y_3&y_4\\z_1&z_2&z_3&z_4\end{array}\right)}_{\text{Tetrahedron} \ T_t}=f_t \ \underbrace{\left(\begin{array}{rrrr}\cos t&-\sin t&0\\ \sin t&\cos t&0\\0&0&1\end{array}\right)}_{\text{Rotation} \ R_t}\underbrace{\left(\begin{array}{rrrr}1&-1&1&-1\\1&1&-1&-1\\1&-1&-1&1\end{array}\right)}_{\text{Tetrahedron T=ABCD}} \tag{1}$$
with $-\frac{\pi}{2} \le t \le \frac{\pi}{2}$ and shrinking factor $f_t=\tfrac{1}{2 \cos(t)+1}$ in the range $1/3 \le f \le 0$ as desired.
Please note that $(T_0)=(T)$ and $(T_{\pi/2})=(T')$.
One verifies that the columns of the first matrix verify resp. the 4 equations of the planes containing the faces of $(T)$:
$$\begin{cases}-x-y-z&=&1\\x-y+z&=&1\\-x+y+z&=&1\\x+y-z&=&1\end{cases}$$
(in the order: faces $BCD, ACD, ABD, ABC$).
The locus of the image of the first column of matrix T gives parametric equations for any branch of one of the hyperbolic arcs described by the vertices, for example the branch pictured in yellow (and green) on the figures:
$$\begin{cases}x&=&\frac{\cos(t)-\sin(t)}{2 \cos(t)+1}\\y&=&\frac{\cos(t)+\sin(t)}{2 \cos(t)+1}\\z&=&\frac{1}{2 \cos(t)+1}\end{cases} \ \ \ \ \text{for} \ \ \ \ -\frac{\pi}{2} \le t \le \frac{\pi}{2}\tag{3}$$
(either we go in one direction, for example following the yellow path, or on the other, the green path)
This branch represents the intersection of
a cone and a plane with resp. equations:
$$\begin{cases}x^2+y^2&=&2z^2\\x+y+z&=&1 \end{cases}\tag{4}$$

Fig. 1. $(T)$ in red (inscribed in a cube) and $(T')$ in blue. The locus of one of the vertices of $(T_t)$ (with $0 \le t \le \pi/2$) is represented in yellow (the symmetric curve in green with $-\pi/2 \le t \le 0$ represents another possibility).

Fig. 2 : The same as figure 1 from above.
Much more can be said. If I have time, I will add elements to this answer.
A: I choose to put the vertices of the outer tetrahedron at $(\pm 2,0,h)$ and $(0,\pm 2,-h)$, where $h=2\sqrt{2}$.
If the inner tetrahedron is scaled by a factor $f$, then its top two vertices will be at height $hf$, so let's look at how the plane $z=hf$ cuts the outer tetrahedron. At $f=0$ it results in the square with vertices $(\pm 1, \pm 1, 0)$, and at $f=1$ it is the degenerate rectangle $(\pm 2, \pm 0, h)$. The tetrahedron edges are straight lines so we can interpolate linearly between these to find the rectangle $(\pm(1+f), \pm(1-f), hf)$.

The top two vertices of the unrotated scaled inner tetrahedron lie at $(\pm 2f, 0, hf)$. If you rotate the inner tetrahedron around the $z$-axis $x=y=0$ until its vertices hit the outer tetrahedron, you can look only in the $z=hf$ plane to see vertex at $(2f,0)$ rotating about $(0,0)$ until it hits a long side of the rectangle at $y=\pm (1-f)$. This obviously means the angle is $\sin^{-1} (\frac{1-f}{2f})$.
Using pythagoras, the coordinate of vertex after rotation is $(\pm\sqrt{3f^2+2f-1}, 1-f, hf)$.
As has already been said, these vertex points for varying $f$ lie on the intersection of a cone, namely $x^2+y^2=(2f)^2=(\frac{2z}{h})^2$, and a plane, the face of the tetrahedron $y=1-f=1-\frac{z}{h}$. Since the plane is at a steeper angle than the cone angle, the vertex points trace out a hyperbola.
