What kind of change of basis keeps the matrix of a linear operator unchanged?

Let $$V$$ be a finite-dimensional vector space in which the set $$mathcal B = {e_i}_{iin{1,cdots,dim V}}$$ is a basis. Consider also the dual space $$V^*$$ and the dual basis $$mathcal B^* = {e^i}_{iin{1,cdots,dim V}}$$. Now let $$A:V to V$$ be any linear operator. The matrix elements of this operator are given by
$$A^i_j = e^i(Ae_j)$$
Then the question is: if one makes a change of basis by a (-n invertible) linear operator $$alpha:V to V, mathcal B mapsto mathcal B’$$, what conditions should $$alpha$$ satisfy such that the matrix elements in the new basis are as those in the old, i.e.
$$A^i_j = e^i(Ae_j) = e^{i’}(Ae_{j’}) = A^{i’}_{j’}$$
where $$mathcal B’ = {alpha e_{i} = e_{i’}}_{iin{1,cdots,dim V}}$$ is the new basis and $$mathcal B’^* = {e^{i’}}_{iin{1,cdots,dim V}}$$ is dual of the new basis. Also the superscript denotes the row and subscript denotes the column.

My Attempt:

By the operator $$alpha:V to V$$ the new basis vectors in terms of the old are $$e_{i’} = alpha^j_{i’} e_j$$ where $$alpha^j_{i’}$$ are matrix elements of $$alpha$$ wrt $$mathcal B$$. The dual of the new basis vectors are $$e^{i’} = alpha_j^{i’} e^j$$, where $$alpha_j^{i’}$$ are matrix elements of the inverse of $$alpha$$ wrt $$mathcal B’$$.

The transformed matrix elements are (in general)
$$e^{i’}(A e_{j’}) = alpha_k^{i’} e^k(A alpha^l_{j’} e_l) = alpha_k^{i’} alpha^l_{j’} e^k(A e_l)$$
requiring that the new matrix elements are the same as the old ones means
$$e^{i}(A e_{j}) = alpha_k^{i’} alpha^l_{j’} e^k(A e_l)$$
Now I write the last equation in matrix notation as
$$(A)_mathcal{B} = (alpha^{-1})_mathcal{B’} (A)_mathcal{B} (alpha)_mathcal{B}$$
Now I use $$(alpha^{-1})_mathcal{B’} = (alpha^{-1})_mathcal{B} = (alpha)_mathcal{B}^{-1}$$ (but I’m not sure of it and it is the reason for stating this question in general) which gives
$$(A)_mathcal{B} = (alpha)_mathcal{B}^{-1} (A)_mathcal{B} (alpha)_mathcal{B} qquad iff qquad (alpha)_mathcal{B}(A)_mathcal{B} – (A)_mathcal{B} (alpha)_mathcal{B} = 0$$
Which is to say that in oreder to keep the matrix elements of $$A$$ in the new basis unchanged one has to change the basis by a(-n invertible linear) operator $$alpha$$ that commutes with $$A$$.
$$(alpha, A) = 0$$

The last result has the following implication: If $$mathcal B$$ is a basis then $$alphamathcal B$$ is as well but also are $$alpha^nmathcal B$$, for $$n in mathbb Z$$ and wrt all these bases the matrix of $$A$$ is the same!

It is good now to ask if there is something wrong with the conclusion $$(alpha, A) = 0$$