[proofplan] We reconstruct every Cartan matrix entry directly from the Dynkin diagram. The diagonal entries are forced to be $2$, and for each unordered pair of distinct vertices the absence, multiplicity, and orientation of the edge determine the corresponding two off-diagonal entries. Finally, an isomorphism of Dynkin diagrams is exactly a relabelling of vertices preserving these data, so it acts on the reconstructed matrix by the same permutation of rows and columns. [/proofplan]

[step:Recover the diagonal entries from the Cartan matrix axioms] Let $A = (a_{ij})_{1 \leq i,j \leq n}$ be a finite-type Cartan matrix. By the definition of a Cartan matrix, every diagonal entry satisfies \begin{align*} a_{ii} = 2 \end{align*} for each $1 \leq i \leq n$. Thus the Dynkin diagram does not need additional information to determine the diagonal: every vertex contributes the diagonal entry $2$. [/step]

[step:Recover zero off-diagonal entries from missing edges] Fix distinct indices $i,j \in \{1,\dots,n\}$. By the definition of the Dynkin diagram, vertices $i$ and $j$ are not joined by an edge exactly when \begin{align*} a_{ij}a_{ji} = 0. \end{align*} By the Cartan matrix axiom $a_{ij} = 0 \iff a_{ji} = 0$, this is equivalent to \begin{align*} a_{ij} = a_{ji} = 0. \end{align*} Therefore a missing edge determines both off-diagonal entries for the pair $(i,j)$. [/step]

[step:Recover nonzero off-diagonal entries from edge multiplicity and orientation] Fix distinct indices $i,j \in \{1,\dots,n\}$ joined by an edge in $\Gamma(A)$. Since $A$ is of finite type, the possible products of opposite off-diagonal entries are \begin{align*} a_{ij}a_{ji} \in \{1,2,3\}. \end{align*} Because $a_{ij}$ and $a_{ji}$ are nonpositive integers and neither is zero, both are negative integers. Hence the edge multiplicity determines the unordered pair $\{a_{ij},a_{ji}\}$ as follows: \begin{align*} a_{ij}a_{ji}=1 &\implies \{a_{ij},a_{ji}\}=\{-1,-1\},\\ a_{ij}a_{ji}=2 &\implies \{a_{ij},a_{ji}\}=\{-1,-2\},\\ a_{ij}a_{ji}=3 &\implies \{a_{ij},a_{ji}\}=\{-1,-3\}. \end{align*} In the single-edge case, the two entries are both $-1$, so no orientation is needed. In the double-edge and triple-edge cases, the Dynkin diagram includes an arrow directed toward the shorter simple root, equivalently toward the vertex $i$ for which $|a_{ij}| > |a_{ji}|$. Therefore the arrow distinguishes which of the two negative integers belongs to which ordered entry. Consequently, for every joined pair of distinct vertices, the diagram determines the ordered pair $(a_{ij},a_{ji})$. [/step]

[step:Compare two matrices using a diagram isomorphism] Let $\varphi:\Gamma(A)\to\Gamma(A')$ be an isomorphism of Dynkin diagrams. Since both diagrams have vertex set cardinality $n$, the map $\varphi$ is a bijection of vertices. Write $\sigma$ for the corresponding permutation of $\{1,\dots,n\}$, so that vertex $i$ of $\Gamma(A)$ is sent to vertex $\sigma(i)$ of $\Gamma(A')$. Because $\varphi$ preserves missing edges, edge multiplicities, and arrow directions, the reconstruction rules from the previous steps give the same ordered entries after relabelling. Thus, for all distinct $i,j$, \begin{align*} a'_{\sigma(i)\sigma(j)} = a_{ij}. \end{align*} For $i=j$, both matrices have diagonal entry $2$, so \begin{align*} a'_{\sigma(i)\sigma(i)} = 2 = a_{ii}. \end{align*} Combining the diagonal and off-diagonal cases yields \begin{align*} a'_{\sigma(i)\sigma(j)} = a_{ij} \end{align*} for all $1 \leq i,j \leq n$. Hence $A'$ is obtained from $A$ by the simultaneous permutation of rows and columns determined by $\sigma$, and the Dynkin diagram determines the finite-type Cartan matrix up to such a permutation. [/step]