The orbit of a Jordan block accumulates on the identity

$I\in\overline{\mathcal{O}(J)}$. For $t>0$, let $D_t=\operatorname{diag}(t,1)$. Then
$$D_tJD_t^{-1}=\begin{pmatrix}t&0\\0&1\end{pmatrix}\begin{pmatrix}1&1\\0&1\end{pmatrix}\begin{pmatrix}t^{-1}&0\\0&1\end{pmatrix}=\begin{pmatrix}t&t\\0&1\end{pmatrix}\begin{pmatrix}t^{-1}&0\\0&1\end{pmatrix}=\begin{pmatrix}1&t\\0&1\end{pmatrix}.$$
As $t\to0$ this tends to $I$, and each $D_tJD_t^{-1}$ belongs to $\mathcal{O}(J)$.

$I\notin\mathcal{O}(J)$. For every $P$, $PIP^{-1}=I$, so the orbit of $I$ is $\{I\}$. If $I=PJP^{-1}$ then $J=P^{-1}IP=I$, which is false.

Consequence. The orbit $\mathcal{O}(J)$ is not closed. In terms of group actions: the quotient $M_n/\!\!\sim$ by conjugation is not Hausdorff. Any continuous function constant on conjugacy classes (such as the trace or the determinant) takes the same values on $J$ and $I$ and indeed $\operatorname{tr}J=\operatorname{tr}I=2$, $\det J=\det I=1$ so it cannot distinguish them.

Remark. $J$ and $I$ have the same characteristic polynomial $(X-1)^2$ but different minimal polynomials. Lesson. Rescaling the basis shrinks off-diagonal parts: this is the mechanism behind all orbit closure arguments.