TITLE: Are all conjugacy classes in $\text{GL}_n(\mathbb R)$ path-connected?
QUESTION [16 upvotes]: Suppose $A$ and $B$ are conjugate invertible real $n \times n$-matrices. Does there always exist a path from $A$ to $B$ inside their conjugacy class?

I thought I had an easy proof for odd $n$ which goes as follows, but it was incorrect as pointed out in this answer. To show where my misunderstanding arised, here is the wrong argument.
Suppose there exists a real matrix $P$ such that $B = PAP^{-1}$. By replacing $P$ with $-P$ if necessary, we can assume that $\det P > 0$ (this is what goes wrong in even dimensions, see this question).
Then we have that $P = e^Q$ for some real matrix $Q$ (since the image of the exponential map is the path-component of the identity in $\text{GL}_n(\mathbb R)$). But now the path $$t \mapsto e^{tQ}Ae^{-tQ}$$
is a path connecting $A$ to $PAP^{-1} = B$.

REPLY [3 votes]: Notice that the assertion "since the image of the exponential map is the path-component of the identity in $GLn(R)$" is false.
Firstly, the path-component of the identity in $GLn(R)$ is the set of  matrices with $>0$ determinant. Thus (inside this component) there is a path linking your $P$ and $I$ (in fact, there is nothing to prove).
Secondly, $diag(-1,-2)$ has $>0$ determinant and is not in the image of the real matrices by the exponential map.
EDIT.
Let $SC(A)$ be the conjugacy class of $A\in M_n(\mathbb{R})$. 
*In dimension $2$, $SC(A)$ is non-connected in 2 cases
a. The eigenvalues of $A$ are non-zero conjugate complex; then $SC(A)$ is homeomorphic to a hyperboloid of two sheets.
b. $A$ is non-zero and non-diagonalizable; then $SC(A)$ is homeomorphic to a conical surface with the apex cut off.
*In dimension $4$, we consider the matrices $A_1=diag(U,U)$ where $U=\begin{pmatrix}0&1\\0&0\end{pmatrix}$ and 
$A_2=diag(V,V)$ where $V=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$. Then, using the Aaron's test, we can prove that $SC(A_1),SC(A_2)$ are non-connected (it's easy for $A_1$ and more difficult for $A_2$).
*Assume that we randomly choose $A\in M_n(\mathbb{R})$ -the $(a_{i,j})$ follow iid normal laws- We deduce that follows 
$\textbf{Proposition}.$ When $n\rightarrow +\infty$, the probability that $SC(A)$ is connected tends to $1$.
$\textbf{Proof}$. A random matrix $A$ has distinct complex eigenvalues with probabiity $1$. Then, up to a real change of basis, $A=diag(a_1I_2+b_1V,\cdots,a_pI_2+b_pV,\lambda_1,\cdots,\lambda_q)$, where $2p+q=n$, the $(b_i)$ are non-zero and the $(\lambda_j)$ are real distinct.
Thus $SC(A)$ is connected iff $q\not=0$ (Aaron's test). When $n$ tends to $+\infty$, the mean of the number of real zeroes of a polynomial of degree $n$ is in $\Omega(\sqrt{n})$; we can deduce that the probability that $A$ has, at least, one real eigenvalue tends to $1$ when $n$ tends to $+\infty$. $\square$