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 $(alpha, A) = 0$