Let $V := M_{n\times n}(F)$ be the space of $n\times n$ matrices over a field $F$. As a finite-dimensional vector space, it admitis a finite basis $\{M^{(j)}\}_{j\in [[1,n^2]]}$. Now my question is very general in nature: is there an inner product, symmetric bilinear form (SBF) or Hermitian form (HF) to decompose an arbitrary $M\in V$ over this basis, in such a way that the components of the matrix agree with the components it would have for a given basis of $F^n$?
The question came to mind when I found out that Pauli matrices, together with the $2\times 2$ identity matrix, provide a basis $\{M^{(j)}\}_{j\in[[1,4]]}$ of $M_{2\times2}(\mathbb{C})$, and that an SBF allowing a satisfactory decomposition of a matrix in this space over that basis is
$$\langle M, N\rangle = \frac{1}{2}\operatorname{Tr}(M N)$$
where by satisfactory I mean here that any matrix, expressed in a particular basis of $\mathbb{C}^2$,
$$\begin{pmatrix}a&b\\c&d\end{pmatrix}\in M_{2\times 2}(\mathbb{C})$$
is invariant under the decomposition, i.e.
$$\begin{pmatrix}a&b\\c&d\end{pmatrix} = \sum_j \frac{1}{2}\operatorname{Tr}(M M^{(j)}) M^{(j)}$$
Now this invariance is dependent on the above specific basis, as it can easily be seen by the fact that over the basis $\{e^{(ij)}: e^{(ij)}_{kl} = \delta_{ki}\delta_{lj}\;\;\forall i,j\in \{1,2\}\}$ the decomposition changes the components. The question, then, can be stated as follows: is there an inner product, SBF or HF over $M_{n\times n}(F)$ that for a given basis of this space the entries of a matrix (over a particular basis of $F^n$) remain invariant? In other words, provided a basis $\{M^{(j)}\}_{j\in [[1,n^2]]}$ and a matrix with components $M_{kl}$, how do I build (if possible) one of the aformentionned forms, such that
$$M_{kl} = \sum_j \langle M, M^{(j)}\rangle M^{(j)}_{kl} \;\;\;\forall k,l\in [\![1,n]\!]$$