Does the KMS condition determine a unique state in a finite dimensional setting? I have two variants of the question, but the specific (rather than general) variant is more important to me.
We consider a finite dimensional Hilbert space $\mathcal H$ and a hermitian operator $H$ which determines the usual time-evolution on $\mathcal L(\mathcal H)$ via $\alpha_t(A)=e^{itH}Ae^{-itH}$ (note that $\alpha(A)$ can extends to an entire analytic function $\mathbb C\to\mathcal L(\mathcal H)$ for any $A$ in $\mathcal L(\mathcal H)$).

Is the KMS condition:
  $$\omega\left(A\,\alpha_{i\beta}(B)\right)=\omega(BA)$$
  only satisfied by the state:
  $$\omega(A)=\frac{\mathrm{Tr}(Ae^{-\beta H})}{\mathrm{Tr}(e^{-\beta H})}$$

So the question is asking whether or not the KMS condition uniquely determines a state in.
A more general version would be the following (but if you answer the above I'm sure I can figure out how to extend it myself):
Let a $C^*$ algebra $\mathcal A$ have a faithful representation on $\mathcal L(\mathcal H)$ with $\mathcal H$ a finite dimensional Hilbert space. Does the KMS condition for a given time-evolution again uniquely determine a state?
 A: Consider a state $\omega$ such that 
$$\tag{1}
\omega\left(A\,\alpha_{i\beta}(B)\right)=\omega(BA).
$$
Because we are in the finite-dimensional case, every state is of the form $\text{Tr}(X\cdot)$, with $X$ positive and $\text{Tr}(X)=1$. So, the equality $(1)$ says that, for every pair of matrices $A,B$, 
$$\tag{2}
\text{Tr}(XAe^{-\beta H}Be^{\beta H})=\text{Tr}(XBA).
$$
Taking $B=e^{\beta H} $, we get $\text{Tr}(e^{\beta H}XA)=\text{Tr}(Xe^{\beta H}A)$ for all $A$; by the Lemma below, $$\tag{3}e^{\beta H}Xe^{-\beta H}=X.$$ Now we can rewrite the equality $(2)$ as 
$$\tag{4}
\text{Tr}(e^{\beta H}XAe^{-\beta H}B)=\text{Tr}(AXB).
$$
As this occurs for all $B$ we obtain, again by the Lemma,
$$\tag{5}
e^{\beta H}XAe^{-\beta H}=AX
$$
This can be re-written, using $(3)$, as 
$$
(e^{\beta H}X)A=A(e^{\beta H}X).
$$
So the matrix $e^{\beta H}X$ commutes with all matrices $A$; this forces $e^{\beta H}X=\lambda I$ for some scalar $\lambda$. From $\text{Tr}(X)=1$, we get 
$$
X=\frac{e^{-\beta H}}{\text{Tr}(e^{-\beta H})},
$$
and then
$$
\omega(A)=\frac{\text{Tr}(e^{-\beta H}A)}{\text{Tr}(e^{-\beta H})}.
$$

Lemma. If $\text{Tr}(XA)=\text{Tr}(YA)$ for all $A$, then $X=Y$.
Proof. We have $\text{Tr}((X-Y)A)=0$ for all $A$. Now take $A=(X-Y)^*$ and use the faithfulness of the trace. 
