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The answer is always yes. Indeed the set is path-connected.

Let $C(f)$ denote the companion matrix associated to the monic polynomial $f$. Every matrix $A$ is similar to a matrix in rational canonical form: $$B=C(f_1)\oplus C(f_1 f_2)\oplus\cdots\oplus C(f_1 f_2,\cdots f_k)$$ where here $\oplus$ denotes diagonal sum. Then $m$ is the degree of $f_1 f_2\cdots f_k$. Starting with $B$ deform each $f_i$ into a power of $x$. We get a path from $B$ to $$B'=C(x^{a_1})\oplus C(x^{a_2})\oplus\cdots\oplus C(x^{a_k})$$ inside $E_m$. There's a path from $B'$ in $E_m$ given by $$(1-t)C(x^{a_1})\oplus (1-t)C(x^{a_2})\oplus\cdots\oplus C(x^{a_k})$$ ending at $$B_m=O\oplus C(x^m).$$ Thus there is a path in $E_m$ from $A$ to $UB_mU^{-1}$ where $U$ is a nonsingular matrix. If $\det(U)\ne0$ then there is a path in $GL_n(\mathbf{R})$ from $U$ to $I$ and so a path in $E_m$ from $A$ to $B_m$. If $m< n$ then there is a matrix $V$ of negative determinant with $VB_m V^{-1}=V$ V^{-1}=B_m$ so that we may take $U$ to have positive determinant.

The only case that remains is when $m=n$. In this case $E_m$ contains diagonal matrices with distinct entries, and each of these commutes with a matrix of negative determinant.
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I think the

The answer is always yes. Considering Indeed the set is path-connected.

Let $C(f)$ denote the companion matrix associated to the monicpolynomial $f$. Every matrix $A$ is similar to a matrix in rationalcanonical form, one can :$$B=C(f_1)\oplus C(f_1 f_2)\oplus\cdots\oplus C(f_1 f_2,\cdots f_k)$$where here $\oplus$ denotes diagonal sum. Then $m$ is the degree of$f_1 f_2\cdots f_k$. Starting with $B$ deform a matrix continuously each $f_i$ into a nilpotent matrixwhile maintining the sizes powerof the constituent companion matrices.Now keep the largest companion matrix fixed and scale down the remainingtowards zero. $x$. We get a continuous path from our starting matrix $B$ to$$B'=C(x^{a_1})\oplus C(x^{a_2})\oplus\cdots\oplus C(x^{a_k})$$inside $E_m$. There's a conjugate path from $B'$ in $E_m$ given by$$(1-t)C(x^{a_1})\oplus (1-t)C(x^{a_2})\oplus\cdots\oplus C(x^{a_k})$$$$B_m=O\oplus C(x^m).$$Thus there is a path in $E_m$ from $A$ to the direct sum $B_m$ of the zero UB_mU^{-1}$where $U$ is a nonsingular matrix. If $\det(U)\ne0$ then there isa path in $GL_n(\mathbf{R})$ from $U$ to $I$ and the companion matrixof so a path in $x^m$. E_m$from $A$ to $B_m$. If the conjugacy $m< n$ then there is by a matrix $V$ of negative determinantwith $VB_m V^{-1}=V$ so that we may take $U$ to have positive determinant, then it canbe done via a path.But

The only case that is sufficient as there remains is when $m=n$. In this case $E_m$ containsdiagonal matrices with distinct entries, and each of these commuteswith a matrix of negative determinantmatrix commuting with $B_m$..
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I think the answer is always yes. Considering the rational canonical form, one can deform a matrix continuously into a nilpotent matrix while maintining the sizes of the constituent companion matrices. Now keep the largest companion matrix fixed and scale down the remaining towards zero. We get a continuous path from our starting matrix to a conjugate $A$ to the direct sum $B_m$ of the zero matrix and the companion matrix of $x^m$. If the conjugacy is by a matrix of positive determinant, then it can be done via a path. But that is sufficient as there is a negative determinant matrix commuting with $B_m$.