Are all conjugacy classes in $\text{GL}_n(\mathbb R)$ path-connected?

We can use the counterexample from your other question to answer this one. Let $A=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$. Matrices $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ must have trace $a+d$ equal to $0$ and determinant $ad-bc=-a^2-bc$ equal to $1$. This implies that such a matrix cannot have $b=0$, since $-a^2\leq 0$. Therefore the set of conjugates of $A$ is disjoint union of open sets defined by $b>0$ and $b<0$. Neither of those sets is empty, since one contains $A$ and the other contains $-A$. Thus the conjugacy class of $A$ is disconnected.

This edit comes a bit late, but the way I see it is a bit different than the other answers so I'll write it anyway.

Again, I use the example from the question that you link to: $A$ is the rotation by $\frac \pi 2$ in the euclidean plane, and $B$ is the rotation by $-\frac \pi 2$.

Now any conjugate of $A$ by a matrix with positive determinant will correspond to a linear map $\varphi$ such that for any non-zero vector $v$, the vectors $v$ and $\varphi (v)$ in this order make a positive basis of the plane.

Conversely, a conjugate of $A$ by a matrix with negative determinant (which is the case of $B$) will correspond to a linear map $\varphi$ such that for any non-zero vector $v$, the vectors $v$ and $\varphi (v)$ in this order make a negative basis of the plane.

A path from $A$ to $B$ has to cross the set of matrices that have real eigenvalues, and such a matrix cannot be conjugate to $A$.

Other people have given counterexamples, so I would like to demonstrate that counterexamples are somewhat rare. Here is an attempt at a conceptual explanation for both why this is not true in general and also when it is true. First, we note that if we have a path $P(t)$ in $GL_n(\mathbb R)$ with $P(0)=P_0$ and $P(1)=P_1$, then $P(t)AP(t)^{-1}$ gives us a path inside the conjugacy class of $A$. Since $GL_n(\mathbb R)$ has two path components (given by sign of determinant), this shows that the conjugacy class of $A$ is the union of two path connected subsets (conjugating by things with positive or negative determinant), and it will be path connected if these two subsets intersect.

If $PAP^{-1}=QAQ^{-1}$ where $\det(P)>0$ and $\det(Q)<0$, we have $A=(P^{-1}Q)A(P^{-1}Q)^{-1}$, so $A$ commutes with a matrix with negative determinant. The converse is also true, so we have the following result.

Lemma: The conjugacy class of $A$ is path connected if and only if $A$ commutes with some matrix with negative determinant.

This gives several conditions that would ensure conjugacy classes are path connected.

• If $n$ is odd, since then $\det(-I_n)=-1$
• If $\det(A)<0$
• If $\mathbb R^n$ as a direct sum of two $A$-invariant subspaces where one summand is odd dimensional, (e.g, if the Jordan normal form of $A$ has a block of odd size with a real eigenvalue).
• If $\mathbb R^{n}$ is the direct sum of two $A$-invariant subspaces, and the restriction of $A$ to either subspace has negative determinant.

One can check that Jordan blocks only commute with matrices that are upper triangular and have only a single eigenvalue, and if $n$ is even and that eigenvalue is real, such a matrix could not have negative determinant. So these give counterexamples.

I suspect one could give a nice characterization of all counterexamples in terms of JNF. However, I have not worked out the details.