algebraic extension ${{\mathbb}{L}}\supseteq{{\mathbb}{K}}$ (but of limited degree).

### Conjugacy Problem {#conjugacy-problem .unnumbered}

When [`IP1S`]{}is given in canonical form, equations  can be rewritten as $A^{\operatorname{T}}\,D\, A = D$ with $D=\operatorname{Diag}(d_1,\ldots,d_n)$ and $H_i^{\prime}=A^{\operatorname{T}}\cdot H_i \cdot A=D\,A^{-1}\,D^{-1}\cdot H_i \cdot
A$ for all $i, 2 \leq i \leq m$. Our task is now to solve the following problem:

Let ${{\mathbb}{K}}^{n\times n}$ be the set of $n\times n$ matrices with entries in ${{\mathbb}{K}}$. Let $\{H_1,\ldots,H_m\}$ and $\{H^{\prime}_1,\ldots,H^{\prime}_m\}$ be two families of matrices in ${{\mathbb}{K}}^{n\times n}$. The [$D$-`OSMC`]{}problem is the task to recover – if any – a $D$-orthogonal matrix $X \in {{\mathbb}{L}}^{n\times n}$, *i.e.* $X^{\operatorname{T}}DX=D$, with ${{\mathbb}{L}}$ being an algebraic extension of ${{\mathbb}{K}}$, such that: $$\begin{aligned}
    X^{-1}\, H_i \, X=H^{\prime}_i, & \quad \forall\,i, 1 \leq i
    \leq m,\end{aligned}$$

[@DBLP:conf/issac/ChistovIK97] show that [$D$-`OSMC`]{}with $D={{\mathrm}{Id}}$ is equivalent to:

1.  Solving the Simultaneous Matrix Conjugacy problem ([`SMC`]{}) between $\{H_{i}\}_{1 \leq i \leq m}$ and $\{H^{\prime}_i\}_{1 \leq
        i \leq m}$, that is to say finding an invertible matrix $Y \in
      \operatorname{GL}_n({{\mathbb}{K}})$ such that: $$\begin{aligned}
    \label{eq:smc0} Y^{-1}\cdot H_i \cdot Y=H^{\prime}_i
        & \text{ and } & Y^{-1}\cdot H_i^{\operatorname{T}} \cdot Y={H^{\prime}_i}^{\operatorname{T}}
        \quad \forall\,i, 1 \leq i \leq m.
      \end{aligned}$$

2.  Computing the square-root $W$ of the matrix $Z=Y \cdot
      Y^{\operatorname{T}}$. Then, the 