Your question is rather vague to me but I'll give it a shot, hoping I understood your intents correctly. In the CJ-isomorphism one considers the map $\mathfrak C:\mathcal L(\mathbb C^{n\times n})\to \mathbb C^{n^2\times n^2}$ which takes any linear operator $T:\mathbb C^{n\times n}\to \mathbb C^{n\times n}$ and maps it to the $n^2\times n^2$-matrix
$$
\mathfrak C(T):=\sum\nolimits_{j,k=1}^n E_{jk}\otimes T(E_{jk})=\begin{pmatrix} T(E_{11})&T(E_{12})&\ldots&T(E_{1n})\\ T(E_{21})&T(E_{22})&\ddots&\vdots\\ \vdots&\ddots&\ddots&T(E_{(n-1)n})\\ T(E_{n1})&\cdots&T(E_{n(n-1)})&T(E_{nn}) \end{pmatrix}
$$
with $(E_{jk})_{j,k=1}^n$ being the standard basis of $\mathbb C^{n\times n}$ and $\otimes$ the usual Kronecker product. Bilinearity of the tensor product shows that $\mathfrak C$ is linear itself (i.e. $\mathfrak C(\lambda S+T)=\lambda \mathfrak C(S)+\mathfrak C(T)$ for all $\lambda\in\mathbb C$, $S,T\in\mathcal L(\mathbb C^{n\times n})$). Now one readily verifies that the map
$$
\big(\mathfrak C^{-1}(A)\big)(X)=\operatorname{tr}_1\big((X^T\otimes\operatorname{id}_n)A\big)\quad \text{ for any }A\in\mathbb C^{n^2\times n^2}\text{ and all }X\in\mathbb C^{n\times n}
$$
is the inverse of $\mathfrak C$, where $\operatorname{tr}_1$ is the partial trace on the first tensor component. Altogether, this shows the following:
$\mathfrak C$ is an isomorphism between $\mathcal L(\mathbb C^{n\times n})$ and $\mathbb C^{n^2\times n^2}$ (and not "only" between the convex cones of completely positive maps and the positive semi-definite $n^2\times n^2$-matrices)
This should not be surprising, given that $\mathfrak C(T)$ contains the action of $T$ on a basis of $\mathbb C^{n\times n}$ which of course ($\to$ linearity!) uniquely determines every such $T$ so one gets a one-to-one correspondence between said spaces.