Every $4\times 4$ complex Hadamard matrix is equivalent to a matrix belonging to the only maximal affine Hadamard family $F_4^{(1)}(a)=F_4\circ{\rm EXP}\left(i R_{F_4^{(1)}}(a)\right)$ stemming from Fourier matrix $F_4$ [32] $$F_4=\left[\begin{array}{rrrr} 1 & 1& 1& 1\\ 1 & i& -1& -i\\ 1 & -1& 1& -1\\ 1 & -i& -1& i \end{array}\right].$$ So we have $$R_{F_4^{(1)}}(a)=\left[\begin{array}{cccc} \bullet & \bullet & \bullet & \bullet\\ \bullet & a & \bullet & a\\ \bullet & \bullet & \bullet & \bullet\\ \bullet & a & \bullet & a \end{array}\right]\Longrightarrow F_4^{(0)}(a)=\left[\begin{array}{rrrr} 1& 1 & 1& 1\\ 1& i\exp(ia)& -1& -i\exp(ia)\\ 1& -1 & 1& -1\\ 1& -i\exp(ia)& -1& i\exp(ia) \end{array}\right].$$
By construction $F_4(0) = F_4$ while $F_4(\pi/2) \simeq F_2 \otimes F_2\not\simeq F_4$.