How to show that $D^{-1}A$ and $L^{T}D^{-1}L$ have same Eigenvalues

Let $A$ be a symmetric positive definite matrix, I want to prove that $D^{-1}A$ and $L^{T}D^{-1}L$ have same Eigenvalues where $D=\text{diag}(\text{diag}(A))$ and $L$ is a lower-triangular matrix such that $A=LL^{T}$.

I observed that $D^{-1}A=D^{-1}LL^{T}$ and thus we can write $L^{T}D^{-1}A=(L^{T}D^{-1}L)L^{T}$.How can we proceed from here? I would hope for some hints. I noticed that $D^{-1}$ and $L^{T}D^{-1}L$ are similar given this form. Therefore, they both must have same eigenvalues but how can I show that $D^{-1}$ and $D^{-1}A$ have same eigenvalues to complete this proof by substitution as it appears?

Update: As mentioned in the comments, this does not work since $L^{T}D^{-1}L$ and $D^{-1}$ are congruent and not similar. Therefore, I would really hope for some help in finding an approach to prove the claim.

Remark: $D:=\text{diag}(\text{diag}(A))$ means that $D$ is a diagonal matrix whose diagonal entries are the diagonal entries of $A$. This is based on MATLAB's notation.

$$D^{-1}L L^t = (L^t)^{-1}( L^t D^{-1} L) L^t.$$ So, the two matrices are conjugate (note that $L$ is invertible because $A$ is positive definite).