= Solution
Define the <change-of-basis matrices> by expressing each new <basis vector> in the corresponding old <basis>:
$$
\mathbf b'_i=\sum_{k=1}^n B_{ki}\mathbf b_k,
\qquad
\mathbf c'_j=\sum_{\ell=1}^m C_{\ell j}\mathbf c_\ell.
$$
Thus $B$ is $n\times n$, $C$ is $m\times m$, and both are <invertible matrices>: a <linear combination> of their columns is zero exactly when the same combination of the new <basis vectors> is zero. For any vector $v$, its coordinate columns therefore satisfy $[v]_{\mathbf b}=B[v]_{\mathbf b'}$; likewise $[w]_{\mathbf c}=C[w]_{\mathbf c'}$.
The <matrix representation of a linear map> gives
$$
C[\Phi(v)]_{\mathbf c'}=[\Phi(v)]_{\mathbf c}
=A[v]_{\mathbf b}=AB[v]_{\mathbf b'}.
$$
Multiplying by the <matrix inverse> of $C$ and comparing with $[\Phi(v)]_{\mathbf c'}=A'[v]_{\mathbf b'}$ for every $v$ proves
$$
\boxed{A'=C^{-1}AB.}
$$
The domain <change of basis> acts on the right, while the codomain <change of basis> acts on the left through its inverse.
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