It is well-known that every square matrix is conjugate to its transpose. This means (in the case of real matrices) that, for each $n\times n$ matrix $M$ with real entries, there is a matrix $S_M\in GL(n,\mathbb{R})$ such that ${S_M}^{-1}MS_M=M^T$. My question is: can you choose $S_M$ in such a way that it depends continuously on $M$? In other words:

Is there a continuous map $\psi\colon M_{n,n}(\mathbb{R})\longrightarrow GL(n,\mathbb{R})$ such that$$\bigl(\forall M\in M_{n\times n}(\mathbb{R})\bigr):\psi(M)^{-1}.M.\psi(M)=M^T?$$

My guess is that the answer is negative even for $n=2$.

Note that, for each individual matrix $M$, there are plenty of choices for $S_M$. For instance, if $n=2$ and$$M=\begin{bmatrix}x&y\\z&t\end{bmatrix},$$then you can take$$S_M=\begin{bmatrix}az&bz\\bz&bt-bx+ay\end{bmatrix},$$with $a$ and $b$ chosen such that $\det(S_M)\neq0$ but, of course, this will only work if $z\neq0$. What if $z=0$? Then you can take$$S_M=\begin{bmatrix}-at+ax&ay\\ay&by\end{bmatrix}$$and, again, $a$ and $b$ should be chosen such that $\det(S_M)\neq0$; the problem now is that, of course, this will only work if $y\neq0$. And so on. This looks like the problem of finding a logarithm for each $z\in\mathbb{C}\setminus\{0\}$: there are plenty of choices for each individual $z$, but there is no continuous way of picking one.