Do these properties hold for trace class operators?

I want to check two properties of trace class operators which I believe to be true an even have some skectches of proofs. In what follows, $$A: D(A) \to \mathscr{H}$$ is a densely defined self-adjoint operator on a Hilbert space $$\mathscr{H}$$.

Suppose $$e^{-tA}$$ is trace class, $$t > 0$$. Then, it is compact, by the spectral theorem, also self-adjoint. Thus, there is a complete orthonormal set $$\{\psi_{n}\}_{n\in \mathbb{N}}$$ composed of eigenvectors of $$e^{-tA}$$ with eigenvalues, say, $$e^{-tA}\psi_{n} = \lambda_{n}\psi_{n}$$. Because $$e^{-tA}$$ is trace class and self-adjoint, its eigenvalues are all positive and satisfy $$\lambda_{n} \to 0$$ as $$n \to \infty$$.

Question 1: Does it follow that $$\{\psi_{n}\}_{n\in \mathbb{N}}$$ is also a set of eigenvectors of $$A$$, with eigenvalues $$\mu_{n} = -\frac{1}{t}\ln\lambda_{n}$$?

I believe the answer is yes, and my reasoning is the following. By the spectral theorem, every eigenvector of $$A$$ is also an eigenvector of $$e^{-tA}$$ and if $$\mu_{n}$$ is an eigenvalue of $$A$$ its associated eigenvalue of $$e^{-tA}$$ will be $$\lambda_{n} = e^{-t\mu_{n}}$$. Conversely, if $$e^{-tA}$$ has an eigenvalue $$\lambda_{n} \ge 0$$ with associated eigenvector $$\psi_{n}$$, then $$\mu_{n} =-\frac{1}{t}\ln\lambda_{n}$$ must be an eigenvalue of $$A$$ with eigenvector $$\psi_{n}$$, since by the spectral theorem $$e^{-tA}\psi_{n} = e^{-t\mu_{n}}$$.

Is my proof correct?

For the second question, suppose $$A$$ has a countable set of eigenvalues $$\{\mu_{n}\}_{n\in \mathbb{N}}$$ and a trace class operator $$B$$ has a set of eigenvalues $$\{e^{-t\mu_{n}}\}_{n\in \mathbb{N}}$$. Question 2: Does it follow that $$B = e^{-tA}$$?

I don't think your proof works. For starters, there is a priori no reason for $$A$$ to have eigenvalues/eigenvectors at all. Also, it is true that $$e^{-tA}$$ is positive, but that's not just because it is selfadjoint, but rather because (as you say) its eigenvalues are of the form $$e^{-t\lambda_n}$$ with $$\lambda_n$$ real; alternatively, $$e^{-tA}$$ is positive because it has a selfadjoint square root.
But the most important problem is that because $$A$$ is densely defined, you have no guarantee whatsoever that $$A\psi$$ exists if $$e^{-tA}\psi=\lambda\psi$$.