I only proved that the diagonals of the bigger quadrilateral meet at F and they are perpendicular. After that ded..

After dual transformation with center $F$ ellipse becomes a circle, intersections of ellipse with perpendicular lines become sides of square, circumscribed around circle. Touch points of square and circle form a square circumscribed around circle, and clearly this circle is imagine of desired conic.

What a pretty problem! Took me a while to get through, but I enjoyed the process and it taught me a lot! Let the ellipse be $\Gamma$. Let $X, Y, Z, W$ be the intersection of the perpendicular lines through $F$ with $\Gamma$, in that order. Let $A = XX\cap WW, B = XX\cap YY, C = YY\cap ZZ, D = ZZ\cap WW$. Let $AC$ intersect $\Gamma$ at $I, K$, with $I$ closer to $A$, and $BD$ intersect the $\Gamma$ at $J, L$, with $J$ closer to $B$. Let $T = XY\cap WZ$. First, observe that Pascal's on $XXZWWY$ gives $A, F, T$ collinear. Similarly, $C, F, T$ collinear, so $T,A,I,F,K,C$ collinear. Similarly, $B,J,F,L, D$ collinear. Since $F$ is the foci of an in-ellipse, this means $\angle BFA + \angle CFD = 90$, but $\angle BFA = \angle CFD$ so $BD\perp CA$. Now, since the pole of $T$ is $BD$, this means $TJ, TL$ are tangent to the $\Gamma$, so $(IK;JL) = -1$; since $XZ\perp YW$, we also have $(XZ;YW) = -1$. Denote $X' = II\cap JJ, Y' = JJ\cap KK, Z' = KK\cap LL, W' = LL\cap II$. We similarly see that $X',X,F,Z,Z'$ and $W',W,F,Y,Y'$ are collinear. Now, because $LL, JJ, WZ$ are concurrent at $T$, this means \[-1 = (WZ;LJ) \overset{L}{=} (WZ;T LF\cap WZ) \overset{D}{=} (AC;TF)\]If we let $S = II\cap KK\cap XW \cap YZ$, and $R = YY\cap WW \cap IL\cap JK$, and $P = XX\cap ZZ \cap IJ \cap KL$, then observe that $T,S,R,P$ lie on $AC, BD, X'Z', Y'W'$ respectively. Furthermore, we have \[-1 = (TF;IK) = (SF;JD) = (RF;XZ) = (PF;YW)\]so $T,S,R,P$ lie on the directrix of $\Gamma$, and are thus collinear. Now, consider the ellipse that goes through $AX'BCZ'$. Because $(AC;FT) = (X'Z';FR) = -1$, this means the pole of $F$ is the directrix, so on a similar note, because $(BD;FS) = -1$, this means $(AX'BCZ'D)$ lies on an ellipse. Similarly, $(ABY'CXW')$ lies on an ellipse; because $F$ has the same pole in both, this means $(AX'BY'CZ'DW')$ lies on an ellipse; let this ellipse be $\Omega$. Now, consider the quadrilateral formed by the tangents from $A,B,C,D$ with $\Omega$. The vertex $A* = AA\cap BB$ lies on the pole of $P$, and so it is collinear with $X,F,Z$. Similarly, $C* = CC\cap DD$ lies on $X,F,Z$, and $B* = BB\cap CC$ and $D* = DD\cap AA$ are collinear with $F$ and $Y$ and $W$. This means $A*C*\perp B*D*$, so $\angle A*FB* + \angle C*FD* = 180$, and $F$ is the foci of some in-ellipse of $A*B*C*D*$, which must be $\Omega$. Thus, $\Omega$ has foci $F$.