Recall that the symplectic group $$Sp_2(\mathbb{R}):= \{A\in SL_2(\mathbb{R}):A^TJA=J\}, \ \ J= \left[ {\begin{array}{cc} 0 & 1 \\ -1 & 0 \\ \end{array} } \right] \ $$ We have its Lie-algebra $\mathfrak{sp}_2(\mathbb{R}):=\{\zeta :Tr(\zeta)=0\}.$
Consider the action by conjugation: $$Sp_2(\mathbb{R}) \times \mathfrak{sp}_2(\mathbb{R}) \to \mathfrak{sp}_2(\mathbb{R}), \ (g,A)\mapsto gAg^{-1}.$$
I need to show that
The spectrum of $\zeta$ is essentially the only invariant of this action.
The way I understood this was to show that if $f:\mathfrak{sp}_2(\mathbb{R}) \to \mathbb{R}$ is a function satisfying $$f(gAg^{-1})=f(A),$$ then either $f$ is a trace function or determinant function. Is this correct way to rephrase my question. If yes, how should I go about proving the claim?
I look forward for suggestions!
First of all, observe that $\textrm{Sp}_2(\mathbb{R})=\textrm{SL}(2,\mathbb{R})$. The symplectic group differs from the special linear group only in dimension $4$ and up. Similarly for the Lie algebra. This will simplify your computation enormously.
Secondly, since you just have $2\times 2$ matrices of zero trace (and at most two distinct eigenvalues), I believe the fastest way to prove your claim is to pick any two matrices $A,B$ with the same spectrum and show that there is a matrix $g$ with real entries of determinant $1$ such that $gAg^{-1}=B$.
This is not hard because 1) trace zero and 2) at most two distinct eigenvalues means you have few computations to do: your general matrix is $$M=\begin{bmatrix}a & b\\c & -a\end{bmatrix}$$ with arbitrary $\lambda$. A representative of matrices with distinct eigenvalues $\lambda\neq -\lambda$ is $$\begin{bmatrix}\lambda & 0\\0 & -\lambda\end{bmatrix}$$ and one with equal eigenvalues is $$\begin{bmatrix}0& 1\\0 & 0\end{bmatrix}.$$ Can you find $g$ that conjugates your $M$ to one of these classes? It is just four equations to solve. Then transitivity of the equivalence under conjugation will finish the problem.
Edit: I forgot the purely imaginary eigenvalues, since we are taking real matrices. You have a third representative for the matrices of purely imaginary eigenvalues $\pm i\lambda$: $$\begin{bmatrix}0 & \lambda \\ -\lambda & 0\end{bmatrix}.$$ Now you can finish mapping everything to one of these three classes of representatives and you will be done.