Suppose that a matrix $M\in\mathrm{SL}_2(\mathbb{C})\setminus\{\pm I\}$ has diagonal elements complex conjugates of each other and that their product is less than $1$. That is, $$M=\begin{pmatrix} a & b\\ c & \bar{a}\end{pmatrix},$$ with $|a|^2<1$.
Also, suppose that there exists a matrix $T\in\mathrm{SL}_2(\mathbb{C})$ such that $T^{-1}MT\in\mathrm{SU}_2$.
Then, I think the following statement holds:
$T$ must be diagonal, that is, $T=\mathrm{diag}(p, 1/p)$.
I would like to prove this statement and have tried the following:
Since there exists a matrix $T\in\mathrm{SL}_2(\mathbb{C})$ such that $T^{-1}MT\in\mathrm{SU}_2$, then $\mathrm{tr}M\in (-2,2)$. The latter is easy to prove using the properties of the trace and those of unitary matrices. Therefore, we can deduce that $\mathrm{Re}(a)\in (-1,1)$. Also, since $|a|^2<1$ and the determinants are equal to $1$, we obtain that $bc<0$.
I have also written (using software) the matrix $T^{-1}MT$ in a very simplified way, using all the previous assumptions.
Still, none of this seems to lead to a conclusion regarding the elements of the matrix $T$.
Any idea or clue is very welcome!
Suppose that $M$ is unitary. Then, for any unitary $T$, $T^{-1}MT$ is a unitary. So any $M,T\in\mathrm{SU}_2$ will do. For instance, take $$ M=T=\begin{bmatrix} 1/2& i\sqrt3/2 \\ i\sqrt3/2& 1/2\end{bmatrix}. $$ Then $T^{-1}MT=M\in \mathrm{SU}_2$, and $T$ is not diagonal. Or, to make a slightly more dramatic example, let $T$ as above and $$M=\begin{bmatrix} 0&1\\-1&0\end{bmatrix}.$$ Then $$ T^{-1}MT=\begin{bmatrix} 1/2& -i\sqrt3/2 \\ -i\sqrt3/2& 1/2\end{bmatrix}\begin{bmatrix} 0&1\\-1&0\end{bmatrix}\begin{bmatrix} 1/2& i\sqrt3/2 \\ i\sqrt3/2& 1/2\end{bmatrix} =\begin{bmatrix} i\sqrt3/2&-1/2 \\ 1/2&-i\sqrt3/2\end{bmatrix}\in\mathrm{SU}_2. $$