Change of basis = similarity?

An $n\times n$ matrix is invertible if and only if the columns are a basis for $\mathbb{R}^n$. So you can view $S^{-1}$ as the change-of-basis matrix that translates from the basis whose vectors are the columns of $S^{-1}$ to the standard basis. Then $S$ is the matrix that translates from the standard basis to the basis whose vectors are the columns of $S^{-1}$.

So when you perform $SAS^{-1}$ you can view it as follows: you give it a coordinate vector in terms of the basis $\beta$ made up of the columns of $S^{-1}$. Then $S^{-1}$ translates this into the standard basis; then you apply $A$ as usual; then you apply $S$ and translate it back into the basis $\beta$. So you can view $SAS^{-1}$ as performing $A$, but in terms of the basis $\beta$.