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The relevant uniqueness is uniqueness of the node set, with $n\ge 1$. Reordering the same nodes cannot change a constraint imposed at every node, so an unsorted “different sequence” obtained only by permutation would not give the claimed counterexample. Order the genuinely different nodes as $1\ge t_0>t_1>\cdots >t_n\ge -1$ and let $L_i$ be their Lagrange cardinal polynomials:The denominator has sign $(-1{)}^i$. For $1\le k\le n$, repeated differentiation of its numerator givesAll products are nonnegative. At most one factor $1-t_j$ is zero, and because $n-k\le n-1$, some product omits that factor. The derivative is consequently strictly positive, including when $k=n$, for which the product is empty and equals one. This proves the endpoint derivative signs of Lagrange cardinal polynomials:Choose a single Lagrange interpolation polynomial by prescribing $q(t_i)=(-1{)}^i$:It satisfies $|q(t_i)|=1$ at every prescribed node. Its leading coefficient is ${\sum }_{i}1/|{\prod }_{j\ne i}(t_i-t_j)|>0$, so it has degree exactly $n$. The endpoint sign formula gives $q^{(k)}(1)={\sum }_i|L_i^{(k)}(1)|$ simultaneously for all $k$.

Interpolate the Chebyshev polynomial $T_n$ on the same nodes. Since $|T_n(t_i)|\le 1$,The only points at which $|T_n|=1$ are its $n+1$ extrema. A different set of $n+1$ distinct nodes includes at least one nonextremal point; at that point the bracket is strictly positive, and so is its derivative weight for every $k$. Moreover $T_n^{(k)}(1)>0$: differentiating its differential equation gives $T_n^{(k+1)}(1)=(n^2-k^2)T_n^{(k)}(1)/(2k+1)$, beginning with $T_n(1)=1$ and ending at $k=n-1$. ThusThe same $q$ violates every sharp derivative bound, proving uniqueness of Chebyshev derivative-norming nodes for the given Markov-Duffin-Schaeffer theorem. The result concerns node sets or consistently ordered sequences; a mere permutation is the literal ordering exception noted above.

For $n\ge 1$, any distinct node set in $[-1,1]$ different from the $n+1$ Chebyshev polynomial extrema admits one degree-$n$ polynomial $q$, bounded by one at all those nodes, whose every nonconstant derivative violates the corresponding sharp endpoint bound. In decreasing order set $q(t_i)=(-1{)}^i$. The endpoint derivative signs of Lagrange cardinal polynomials give $q^{(k)}(1)={\sum }_i|L_i^{(k)}(1)|$. Interpolating $T_n$ on the same nodes showsEach term is nonnegative and at least one is positive, because the other node set includes a nonextremal point. Uniqueness is a claim about node sets: permuting the same nodes does not change their norming property.