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What is a matrix?
How do you add or subtract a matrix?
How do you multiply two matrices together?
Let $M_{u,v}$ be the base change matrix from base $u$ to base $v$. Then the following identity holds for bases $u$,$v$, and $w$: $$ M_{u,v}=M_{w,v}M_{u,w}$$ In this case $A$ can be the role of $u$ and $B$ is $v$. Use the standard basis $e$ as the intermediate basis $w$. Then the expression is: $$M_{A,B}=M_{e,B}M_{A,e} = M_{B,e}^{-1}M_{A,e}$$ The basis change matrix from any basis to the standard basis is easy, just write the vectors as the columns of the matrix. So here have to calculate $$\left( \begin{array}{ccc} 1 & -2 & 1 \\ 3 & -1 & 2 \\ 2 & 1 & 3 \end{array} \right)^{-1}\left( \begin{array}{ccc} 1 & 4 & 1 \\ 0 & 5 & 1 \\ 5 & 5 & 4 \end{array} \right)$$
Step 1: Find a change of basis matrix from $A$ to the standard basis Step 2: Do the same for $B$ Step 3: Apply the first, then the inverse of the second.
For the first, if have the coordinates $(p, q, r)$ in the $A$ basis, then in the standard basis, you have $\begin{pmatrix}{1\\0\\5}\end{pmatrix}p + \begin{pmatrix}{4\\5\\5}\end{pmatrix}q + \begin{pmatrix}{1\\1\\4}\end{pmatrix}r $. That means that the change of basis matrix (from the $A$-basis to the standard basis) is exactly what you've called $A$. Similarly for part 2. So the thing you want is $B^{-1} A$. You'll need to invert the matrix $B$, but I'm assuming you can do that.
High level lesson: reduce to a simpler problem when possible.
Without saying much, here is how I usually remember the statement and also the proof in one big picture:
\begin{array}{ccc} x_{1},\dots,x_{n} & \underrightarrow{\;\;\; A\;\;\;} & Ax_{1},\dots,Ax_{n}\\ \\ \uparrow F & & \downarrow F^{-1}\\ \\ y_{1},\dots,y_{n} & \underrightarrow{\;\;\; B\;\;\;} & By_{1},\dots,By_{n} \end{array}
And $$By=F^{-1}AFy$$
Your approach seems correct.
I don't know if the following helps, but anyway: So you have a vector space $V$ (over say the complex numbers) and you have say two basis $E$ and $D$.
That $F$ is a change of basis matrix means that if as column vector $y = (y_i)$ written with respect to the basis $E$, then you get the coordinates with respect to $D$ by $x = Fy$.
Now you have a linear transformation $T: V \to V$. With respect to each basis, this transformation is given by two matrices, say $A_E$, $A_D$. So if $y = (y_i)$ (wrt. basis $E$) then $Ty = A_Ey$ and the result with be the coordinates in the basis $E$. (And likewise for the basis $D$ using $A_D$).
So given a vector $y = (y_i)$ written in the basis $E$, you could then first transform the coordinates to the basis $D$, then you the matrix $A_D$ and then transform the coordinated back to the basis $E$. So you get $A_E(y) = F^{-1}A_DFy$.
One can actually write out all of this (If you have never done so I recommend that you do it) with coordinates. So you would start with the vector $v$ and write it as a linear combination of the basis $E$: $v = y_1e_1 + \dots y_ne_n$ and continue from there...