Characterizing duals of cones that are linear images of the positive semidefinite cone Let $M_n$ denote the space of $n\times n$ matrices over complex numbers. The space of self-adjoint matrices is denoted 
$$
M_n^{sa} = \{A\in M_n\, :\, A^*=A \},
$$
where $A^*$ denotes the conjugate transpose of $A$, 
and the cone of positive semidefinite matrices will be denoted 
$$
 M_n^+ = \{A\in M_n^{sa}\, :\, A\geq0\},
$$
where $A\geq0$ denotes that $A$ is positive semidefinite.
The adjoint of a linear map of matrices $\phi:M_n\rightarrow M_m$ is the unique map $\phi^*:M_m\rightarrow M_n$ that satisfies 
$$
\mathrm{Tr}(\phi(A)B) = \mathrm{Tr}(A\phi^*(B))
$$
for all $A\in M_n$ and $B\in M_m$. The dual cone of a 
subset $S\subseteq M_{n}^{sa}$ is defined by
$$
S^* = \{B\in M_{n}^{sa}\, : \, \forall A\in S, \, \mathrm{Tr}(AB)\geq 0\}
$$
Note that $S^*$ is always a closed and convex cone. 
Let $\phi$ be a linear map of matrices and consider the cones of self-adjoint matrices defined by
$$
K = \{A\in M_n^{sa} \, :\, \phi(A)\geq 0\}
$$
and 
$$
L= \{\phi^*(B) \, :\, B\geq 0\}.
$$
It is evident that both $K$ and $L$ are both convex cones and that $L\subseteq K^*$ and $K\subseteq L^*$. 

Question: Is it necessarily the case that $K^*=L$ or $L^*=K$? If not, under what circumstances would it hold that $K^*=L$ or $L^*=K$? For the linear maps $\phi$ that I've played around with so far this seems to be the case and I'm wondering if there is a counterexample.

 A: We have $K = \phi^{-1}(M_n^+)$ and $L = \phi^*(M_n^+),$ where $\phi^{-1}(S)$ is notation for $\{x : \phi(x) \in S\}$ whether $\phi$ is invertible or not. So the question is whether or not $\phi^*(M_n^+)^* = \phi^{-1}(M_n^+)$ and $\phi^{-1}(M_n^+)^* = \phi^*(M_n^+)$
It's easier to keep track of what to expect when working in more generality, so consider a linear map $\phi:V \to W$, where $V$ and $W$ are finite-dimensional inner product spaces over $\mathbb R$. If you're comfortable with dual spaces, you could work in even more generality and drop the inner product structure.
Given a subset $S \subseteq W$, it's not sensible to expect a nice relationship in general between $\phi^{-1}(S)^*$ and $\phi^*(S)$. For example, adding elements to $S$ makes one of those sets smaller but the other one larger.
But it is sensible to expect one between $\phi^{-1}(S^*)$ and $\phi^*(S)^*$. Indeed,
$$
\phi^*(S)^*
= \{v \in V : \langle v, \phi^*(s)\rangle \geq 0\;\forall s\in S \}
= \{v \in V : \langle \phi(v), s\rangle \geq 0\;\forall s\in S \}
= \phi^{-1}(S^*).
$$
In your situation, we have $V = W = M_n^{sa}$ with inner product $\langle A, B \rangle = \operatorname{Tr}(AB)$, and $S = M_n^+$. Note that ${M_n^+}^* = M_n^+$. So we have
$$L^* = \phi^*(M_n^+)^* = \phi^{-1}({M_n^+}^*) = \phi^{-1}(M_n^+) = K.$$
From this point of view, the missing ingredient is the observation that ${M_n^+}^* = M_n^+$, not just ${M_n^+}^* \supseteq M_n^+$.
It follows that $K^* = L^{**} = \bar L,$ so $K^* = L$ if and only if $L$ is closed, which isn't always the case, e.g. the counterexample in the comments.
