Why does $A^{-1}MA$ give a vector that is in the "language" of basis $A$? For context, I am struggling with understanding the end of this video.
Let's say we have the following 4 sets of basis vectors:
$$B_1 =\begin{bmatrix}
1 & 0 \\
0 & 1  
\end{bmatrix}, B_2 =\begin{bmatrix}
2 & -1 \\
1 & 1  
\end{bmatrix}, B_3 =\begin{bmatrix}
0 & -1 \\
1 & 0  
\end{bmatrix}, B_4 = B_2^{-1} =\begin{bmatrix}
\frac{1}{3} & \frac{1}{3} \\
-\frac{1}{3} & \frac{2}{3}  
\end{bmatrix}$$
Now I want to see what the vector, $v = \begin{bmatrix}
1 \\
0  
\end{bmatrix}$, represents after I apply all the change of basis matrices above on it.

*

*$B_2*v = B_1* \begin{bmatrix}
2 \\
1  
\end{bmatrix}$
What this means: The vector from the linear combination of $1\hat{i}_{B_2} + 0\hat{j}_{B_2}$ is equivalent to the vector $\begin{bmatrix}
2 \\
1  
\end{bmatrix}$ in the basis $B_1$.


*$B_3*\begin{bmatrix}
2 \\
1  
\end{bmatrix} = B_1* \begin{bmatrix}
-1 \\
2  
\end{bmatrix}$
What this means: The vector from the linear combination of $2\hat{i}_{B_3} + 1\hat{j}_{B_3}$ is equivalent to the vector $\begin{bmatrix}
-1 \\
2  
\end{bmatrix}$ in the basis $B_1$.


*$B_4*\begin{bmatrix}
-1 \\
2  
\end{bmatrix} = B_1* \begin{bmatrix}
\frac{1}{3} \\
\frac{5}{3}  
\end{bmatrix}$
What this means: The vector from the linear combination of $-1\hat{i}_{B_4} + 2\hat{j}_{B_4}$ is equivalent to the vector $\begin{bmatrix}
\frac{1}{3} \\
\frac{5}{3}  
\end{bmatrix}$ in the basis $B_1$.
In the video, if I understood correctly, he says that this final vector $\begin{bmatrix}
\frac{1}{3} \\
\frac{5}{3}  
\end{bmatrix}$ is actually in the basis $B_2$, not $B_1$. What am I confounding here?
 A: It's difficult to diagnose properly, but I think the root of your issues comes from confusing bases with invertible matrices. While bases do correspond with invertible matrices, it can confuse your intuition if you treat every invertible matrix as some kind of basis, which appears to be what you've done here.
It's better to think of bases not as matrices, but as sets of vectors. Or, even better, think of them as ordered lists of vectors, as order is important when using bases, and sets don't convey such order. So, if you want the standard basis, think of the ordered list:
$$\mathcal{S} = \left(\begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \end{bmatrix}\right)$$
instead of the identity matrix
$$I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.$$
I know it doesn't seem like much of a distinction, but I honestly think that not making this distinction is at the heart of your problems here.
In particular, you seem to be interpreting $B_4 = B_2^{-1}$ as a basis. I wouldn't recommend this. Jennifer's basis is
$$\mathcal{J} = \left(\begin{bmatrix} 2 \\ 1 \end{bmatrix}, \begin{bmatrix} -1 \\ 1 \end{bmatrix}\right).$$
Now, the change-of-basis matrix from $\mathcal{J}$ to $\mathcal{S}$ is given by
$$B_2 = \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix}.$$
What does this matrix do? If we take a vector $v = \begin{bmatrix} a \\ b \end{bmatrix}$ in Jennifer's "language" (i.e. $a$ times her first basis vector plus $b$ times her second basis vector), then $B_2 v = \begin{bmatrix} c \\ d \end{bmatrix}$ refers to the same point in our language (i.e. $c$ times $(1, 0)$ plus $d$ times $(0, 1)$).
You've already provided a concrete example of this. If we have $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ in her "language", then it refers to $1$ times her first basis vector plus $0$ times her second basis vector. Her first basis vector, in our language, is $\begin{bmatrix} 2 \\ 1 \end{bmatrix}$, and you can verify that
$$\begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}.$$
Importantly, I'm treating $B_2$ not as a basis, but as a multiplicand that helps us translate from one language to another. If Kathrine had a third "language", which was represented by some other basis $\mathcal{K}$, then Jennifer and Kathrine would have a totally different change of basis matrix, whose columns would be neither the basis vectors in $\mathcal{J}$ or $\mathcal{K}$. The matrix would still be invertible, and yes, you could make a basis by taking the columns of this matrix, but that basis would have no intuitive bearing on the problem at hand. That is, it would not be a useful basis for understanding Jennifer or Kathrine.
That's what you're doing wrong with $B_4$. Think of it instead as the reverse process of translating from Jennifer's language into our own. Since it is the inverse of $B_2$, we can multiply it to a vector in our language, and get the same vector in Jennifer's language. If you harvest a new basis $\mathcal{B}$ from its columns, you've invented a new language that has nothing intuitive to do with the existing languages, and $B_4$ becomes the translation from this erroneous language $\mathcal{B}$ to our language $\mathcal{S}$.
I'd also like to point out that you're doing a similar thing with $B_3$. The matrix $B_3$ is the transformation matrix. It's not translating the same vector in different languages (at least, it isn't in the context of our problem), it's actually changing the point we're referring to, by rotating it $90^\circ$. It is an invertible matrix, so a basis can be formed, but we gain nothing, intuitively speaking, by interpreting it as such. Plus, we can just as easily have a non-invertible transformation here (e.g. projecting onto the $x$-axis), in which case a basis cannot be formed from the columns!
So yeah, TL;DR: not every matrix should be interpreted as a basis.
