Similarity is an equivalence relation, and two matrices represent the same endomorphism in two bases exactly when they are similar

Taking $P=I_n$ proves reflexivity. If $B=P^{-1}AP$, then $A=(P^{-1}{)}^{-1}B{P}^{-1}$, proving symmetry. If also $C=Q^{-1}BQ$, then associativity in [L4] gives $C=(PQ{)}^{-1}A(PQ)$, proving transitivity. Thus similarity is an equivalence relation.

If $A$ and $B$ represent the same endomorphism in bases $\mathcal{B}$ and $\mathcal{C}$, [L2] gives $B=P^{-1}AP$ with $P=P_{\mathcal{B}\leftarrow \mathcal{C}}$, so they are similar.

Conversely, suppose $B=P^{-1}AP$. On $V=M_{n\times 1}(F)$ let $T=L_A$, which is linear by [L4], and let $\mathcal{E}=(E_{i0}{)}_{i<n}$. By [L3], every column $y$ has the unique expansion $y={\sum }_{i<n}{y}_i{E}_{i0}$, so [L5] makes $\mathcal{E}$ the standard ordered basis. The columns of invertible $P$ form an ordered basis $\mathcal{C}$: independence follows by multiplying $Px=0$ by $P^{-1}$, and every column vector $y$ equals $P(P^{-1}y)$, so [L5] applies. Moreover, $[T{]}_{\mathcal{E}}^{\mathcal{E}}=A$ because $T(E_{j0})$ is the $j$-th column of $A$. The change matrix $P_{\mathcal{E}\leftarrow \mathcal{C}}$ is $P$, so [L2] gives $[T{]}_{\mathcal{C}}^{\mathcal{C}}=P^{-1}AP=B$.

Steps 2.1 and 3.1 prove both directions of the characterisation, including $n=0$, where the unique empty matrix represents the unique endomorphism of the zero space.