I will be using the set-theoretic notation $0=\emptyset$, $\forall n\in\mathbb{N}^*: n=\{0,1,...,n-1\}$ and $B^A=\{f:A\to B\}$ so the cartesian power $X^n$ can be expressed as $X^n =\{f:n\to X\}$ where we identify $f=(f(0),f(1),...,f(n-1))$. Also, I will denote similar matrices as $A\simeq B\in M_n(\mathbb{C})\iff\exists P\in GL_n(\mathbb{C}):PAP^{-1}=B$. Now, let me first contextualize the question.
The symmetric group $\Sigma_n$ acts on $\mathbb{C}^n$ as $\sigma*v\mapsto v\circ\sigma$ so we can consider the quotient $\mathbb{C}^n/\Sigma_n$ with its quotient topology. We can also endow the set $\text{Mon}(\mathbb{C}_n[z])$ of complex monic polynomials of degree $n$ with the standard topology of $\mathbb{C}^n$ with the natural homeomorphism $z^n+a_{n-1}z^{n-1}+...+a_0\mapsto(a_0,...,a_{n-1})$. Then, the map $\mathbb{C}^n/\Sigma_n\to\text{Mon}(\mathbb{C}_n[z])$ defined as
$$[\lambda_0,...,\lambda_{n-1}]\mapsto\prod_{i\in n}(z-\lambda_i)$$
is in fact a homeomorphism as it has already been discussed here; I will call it Viète's homeomorphism. It is not hard to see via induction that the characteristic polynomial function $\chi:M_n(\mathbb{C})\to\text{Mon}(\mathbb{C}_n[z])$ is continuous by endowing $M_n(\mathbb{C})$ with the standard topology of $\mathbb{C}^{n^2}$ since the coefficients are polynomial functions. In fact, as similar matrices have the same characteristic polynomial, it induces a continuous function $\chi:M_n(\mathbb{C})/\simeq\to\text{Mon}(\mathbb{C}_n[z])$. By restricting the domain to the set $\text{Diag}(M_n(\mathbb{C}))$ of diagonalizable matrices we have that $\chi:\text{Diag}(M_n(\mathbb{C}))/\simeq\to\text{Mon}(\mathbb{C}_n[z])$ is continuous and bijective since two diagonalizable matrices are similar $\iff$ they have the same characteristic polynomial. Moreover, with Viète's homeomorphism, it can be proven that the inverse $\chi^{-1}:\text{Mon}(\mathbb{C}_n[z])\to\text{Diag}(M_n(\mathbb{C}))/\simeq$
$$\prod_{i\in n}(z-\lambda_i)\mapsto[\text{diag}(\lambda_0,\lambda_1,...,\lambda_{n-1})]$$
is also continuous. This is because, if $\nu:\mathbb{C}^n/\Sigma_n\to\text{Mon}(\mathbb{C}_n[z])$ is Viète's homeomorphism and $\mu:\mathbb{C}^n/\Sigma_n\to\text{Diag}(M_n(\mathbb{C}))/\simeq$ is the continuous function $[\lambda_0,...,\lambda_{n-1}]\mapsto[\text{diag}(\lambda_0,...,\lambda_{n-1})]$, then $\chi^{-1}=\mu\circ\nu^{-1}$ is continuous for being the composition of continuous functions. Hence
$$\boxed{\mathbb{C}^n/\Sigma_n\cong \text{Mon}(\mathbb{C}_n[z])\cong\text{Diag}(M_n(\mathbb{C}))/\simeq}$$
With this in mind, the goal is clear: to generalize this for $M_n(\mathbb{C})/\simeq$, not only diagonalizable matrices. Clearly, $\mathbb{C}^n/\Sigma_n$ won't suffice as the eigenvalues with their associated algebraic multiplicity don't characterize the conjugacy class of general matrices. We need a natural way to describe the Jordan blocks with tuples. What I've came up with is the following. Define
$$J(\mathbb{C}^n):=\{(v,f)\in\mathbb{C}^{\subseteq n}\times n^n:\text{Dom}(v)=\text{Im}(f)\}/\sim$$
where $(v_1,f_1)\sim(v_2,f_2)\iff v_1\circ f_1=v_2\circ f_2\land\exists \sigma\in\Sigma_n:\sigma\circ f_1=f_2$ and $\mathbb{C}^{\subseteq n}=\amalg_{S\subseteq n}\mathbb{C}^{S}$. Also, with $n^n$ I mean $n^n=\{f:n\to n\}$. The idea is that any Jordan normal form can be described as, by denoting $J_{n_i}(\lambda_i)$ as the Jordan matrix of size $n_i$ with $\lambda_i$'s on the diagonal, $$J = J_{n_0}(\lambda_0)\oplus J_{n_1}(\lambda_1)\oplus...\oplus J_{n_{k-1}}(\lambda_{k-1})\mapsto v_J = \begin{pmatrix}0 & ... & k-1 \\ \lambda_0 & ... & \lambda_{k-1}\end{pmatrix}$$ $$\text{and }\ f_J = \begin{pmatrix}0 & ... & n_{0}-1& n_0 & ... & n_0+n_1-1 & ... & n-n_{k-1}+1 & ... & n \\ 0 & ... & 0 & 1 & ... & 1 & ... & k-1 & ... & k-1 \end{pmatrix}$$ $$\text{so }\ v_J\circ f_J = \begin{pmatrix}0 & ... & n_{0}-1& n_0 & ... & n_0+n_1-1 & ... & n-n_{k-1}+1 & ... & n \\ \lambda_0 & ... & \lambda_0 & \lambda_1 & ... & \lambda_1 & ... & \lambda_{k-1} & ... & \lambda_{k-1} \end{pmatrix}$$ That is, $v_J\circ f_J$ are the eigenvalues with their respective algebraic multiplicity (a vector in $\mathbb{C}^n$). But, $(v_J,f_J)$ gives us more information than that: $v_J$ tells us which eigenvalues are associated to each Jordan block and $f_J$ tells us which are the sizes of each Jordan block. The $i^{th}$ Jordan block has size $n_i=|f_J^{-1}(\{i\})|$ and eigenvalue $\lambda_i=v_J(i)$. We can endow $J(\mathbb{C}^n)$ with a topology by considering on $\mathbb{C}^{\subseteq n}$ the disjoint union topology (where each $\mathbb{C}^S$ has the standard topology of $\mathbb{C}^{|S|}$) and on $n^n$ some other topology $\mathcal{T}(n^n)$ (maybe the trivial topology?). Then, $\mathbb{C}^n$ is embedded naturally on $J(\mathbb{C}^n)$ with the mapping $\mathbb{C}^n\to J(\mathbb{C}^n),\ z\mapsto [z,id_n]$. Now, for the last step, note that $\Sigma_n$ acts on $J(\mathbb{C}^n)$ as $\sigma*[v,f]\mapsto[v,f\circ\sigma]$ so we can consider $J(\mathbb{C}^n)/\Sigma_n$ with its quotient topology. Now,
$\bf{Q:}$ Does there exist a topology $\mathcal{T}(n^n)$ of $n^n$ (maybe the trivial topology?) that makes the mapping $\Phi:M_n(\mathbb{C})/\simeq\to J(\mathbb{C}^n)/\Sigma_n$ defined as $[A]=[J_A]\mapsto[v_{J_A},f_{J_A}]$ a homeomorphism? Here, $J_A$ is the Jordan normal form of $A$ and $[J]\mapsto[v_J,f_J]$ works as before. That is, in this sense, can we make the Jordan normal form bicontinuous?
