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For the case $n=8$, with the precoloring you describe the completion you give is indeed unique. I checked by writing the corresponding boolean program and let a solver enumerate all solutions: there is only one.

For the case $n=10$, consider the pre-colored $K_{10}$ $$\left(\begin{array}{rrrrrrrrrr} X & & & & & 8 & 4 & 9 & 5 & 1 \\ & X & & & & & 9 & 5 & 1 & 2 \\ & & X & & & & & 1 & 6 & 3 \\ & & & X & & & & & 2 & 4 \\ & & & & X & & & & & 5 \\ 8 & & & & & X & & & & \\ 4 & 9 & & & & & X & & & \\ 9 & 5 & 1 & & & & & X & & \\ 5 & 1 & 6 & 2 & & & & & X & \\ 1 & 2 & 3 & 4 & 5 & & & & & X \end{array}\right)$$ This can be completed in $333$ ways, for example $$\left(\begin{array}{rrrrrrrrrr} X & 6 & 2 & 7 & 3 & 8 & 4 & 9 & 5 & 1 \\ 6 & X & 7 & 8 & 4 & 3 & 9 & 5 & 1 & 2 \\ 2 & 7 & X & 5 & 9 & 4 & 8 & 1 & 6 & 3 \\ 7 & 8 & 5 & X & 6 & 9 & 1 & 3 & 2 & 4 \\ 3 & 4 & 9 & 6 & X & 1 & 2 & 7 & 8 & 5 \\ 8 & 3 & 4 & 9 & 1 & X & 5 & 2 & 7 & 6 \\ 4 & 9 & 8 & 1 & 2 & 5 & X & 6 & 3 & 7 \\ 9 & 5 & 1 & 3 & 7 & 2 & 6 & X & 4 & 8 \\ 5 & 1 & 6 & 2 & 8 & 7 & 3 & 4 & X & 9 \\ 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & X \end{array}\right) $$ or $$ \left(\begin{array}{rrrrrrrrrr} X & 6 & 2 & 7 & 3 & 8 & 4 & 9 & 5 & 1 \\ 6 & X & 4 & 8 & 7 & 3 & 9 & 5 & 1 & 2 \\ 2 & 4 & X & 5 & 8 & 9 & 7 & 1 & 6 & 3 \\ 7 & 8 & 5 & X & 9 & 1 & 6 & 3 & 2 & 4 \\ 3 & 7 & 8 & 9 & X & 2 & 1 & 6 & 4 & 5 \\ 8 & 3 & 9 & 1 & 2 & X & 5 & 4 & 7 & 6 \\ 4 & 9 & 7 & 6 & 1 & 5 & X & 2 & 3 & 8 \\ 9 & 5 & 1 & 3 & 6 & 4 & 2 & X & 8 & 7 \\ 5 & 1 & 6 & 2 & 4 & 7 & 3 & 8 & X & 9 \\ 1 & 2 & 3 & 4 & 5 & 6 & 8 & 7 & 9 & X \end{array}\right)$$

So it looks very plausible to me, that the completion can always be done for $n\geq 8$ and it is not unique for $n\geq 10$.

For the case $n=8$, with the precoloring you describe the completion you give is indeed unique. I checked by writing the corresponding boolean program and let a solver enumerate all solutions: there is only one.

For the case $n=10$, consider the pre-colored $K_{10}$ $$\left(\begin{array}{rrrrrrrrrr} X & & & & & 1 & 8 & 4 & 9 & 5 \\ & X & & & & & 2 & 9 & 5 & 1 \\ & & X & & & & & 3 & 1 & 6 \\ & & & X & & & & & 4 & 2 \\ & & & & X & & & & & 7 \\ 1 & & & & & X & & & & \\ 8 & 2 & & & & & X & & & \\ 4 & 9 & 3 & & & & & X & & \\ 9 & 5 & 1 & 4 & & & & & X & \\ 5 & 1 & 6 & 2 & 7 & & & & & X \end{array}\right)$$ This can be completed in $77$ ways, for example $$\left(\begin{array}{rrrrrrrrrr} X & 6 & 7 & 3 & 2 & 1 & 8 & 4 & 9 & 5 \\ 6 & X & 8 & 7 & 3 & 4 & 2 & 9 & 5 & 1 \\ 7 & 8 & X & 5 & 4 & 2 & 9 & 3 & 1 & 6 \\ 3 & 7 & 5 & X & 9 & 8 & 6 & 1 & 4 & 2 \\ 2 & 3 & 4 & 9 & X & 5 & 1 & 6 & 8 & 7 \\ 1 & 4 & 2 & 8 & 5 & X & 3 & 7 & 6 & 9 \\ 8 & 2 & 9 & 6 & 1 & 3 & X & 5 & 7 & 4 \\ 4 & 9 & 3 & 1 & 6 & 7 & 5 & X & 2 & 8 \\ 9 & 5 & 1 & 4 & 8 & 6 & 7 & 2 & X & 3 \\ 5 & 1 & 6 & 2 & 7 & 9 & 4 & 8 & 3 & X \end{array}\right) $$ or $$\left(\begin{array}{rrrrrrrrrr} X & 7 & 2 & 6 & 3 & 1 & 8 & 4 & 9 & 5 \\ 7 & X & 8 & 3 & 4 & 6 & 2 & 9 & 5 & 1 \\ 2 & 8 & X & 5 & 9 & 4 & 7 & 3 & 1 & 6 \\ 6 & 3 & 5 & X & 8 & 7 & 9 & 1 & 4 & 2 \\ 3 & 4 & 9 & 8 & X & 5 & 1 & 6 & 2 & 7 \\ 1 & 6 & 4 & 7 & 5 & X & 3 & 2 & 8 & 9 \\ 8 & 2 & 7 & 9 & 1 & 3 & X & 5 & 6 & 4 \\ 4 & 9 & 3 & 1 & 6 & 2 & 5 & X & 7 & 8 \\ 9 & 5 & 1 & 4 & 2 & 8 & 6 & 7 & X & 3 \\ 5 & 1 & 6 & 2 & 7 & 9 & 4 & 8 & 3 & X \end{array}\right)$$

This answers your question about uniqueness. So it looks very plausible to me, that the completion can always be done for $n\geq 8$ and it is not unique for $n\geq 10$.

I don't quite get what you mean by "distinct consecutive transversals", can you explain more clearly how the pre-coloring is done? I assume you mean fixing some values that don't contradict a valid coloring for the k outer diagonals.

For the case $n=8$, with the precoloring you describe the completion you give is indeed unique. I checked by writing the corresponding boolean program and let a solver enumerate all solutions: there is only one.

For the case $n=10$, consider the pre-colored $K_{10}$ $$\left(\begin{array}{rrrrrrrrrr} X & & & & & 8 & 4 & 9 & 5 & 1 \\ & X & & & & & 9 & 5 & 1 & 2 \\ & & X & & & & & 1 & 6 & 3 \\ & & & X & & & & & 2 & 4 \\ & & & & X & & & & & 5 \\ 8 & & & & & X & & & & \\ 4 & 9 & & & & & X & & & \\ 9 & 5 & 1 & & & & & X & & \\ 5 & 1 & 6 & 2 & & & & & X & \\ 1 & 2 & 3 & 4 & 5 & & & & & X \end{array}\right)$$ This can be completed in $333$ ways, for example $$\left(\begin{array}{rrrrrrrrrr} X & 6 & 2 & 7 & 3 & 8 & 4 & 9 & 5 & 1 \\ 6 & X & 7 & 8 & 4 & 3 & 9 & 5 & 1 & 2 \\ 2 & 7 & X & 5 & 9 & 4 & 8 & 1 & 6 & 3 \\ 7 & 8 & 5 & X & 6 & 9 & 1 & 3 & 2 & 4 \\ 3 & 4 & 9 & 6 & X & 1 & 2 & 7 & 8 & 5 \\ 8 & 3 & 4 & 9 & 1 & X & 5 & 2 & 7 & 6 \\ 4 & 9 & 8 & 1 & 2 & 5 & X & 6 & 3 & 7 \\ 9 & 5 & 1 & 3 & 7 & 2 & 6 & X & 4 & 8 \\ 5 & 1 & 6 & 2 & 8 & 7 & 3 & 4 & X & 9 \\ 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & X \end{array}\right) $$ or $$ \left(\begin{array}{rrrrrrrrrr} X & 6 & 2 & 7 & 3 & 8 & 4 & 9 & 5 & 1 \\ 6 & X & 4 & 8 & 7 & 3 & 9 & 5 & 1 & 2 \\ 2 & 4 & X & 5 & 8 & 9 & 7 & 1 & 6 & 3 \\ 7 & 8 & 5 & X & 9 & 1 & 6 & 3 & 2 & 4 \\ 3 & 7 & 8 & 9 & X & 2 & 1 & 6 & 4 & 5 \\ 8 & 3 & 9 & 1 & 2 & X & 5 & 4 & 7 & 6 \\ 4 & 9 & 7 & 6 & 1 & 5 & X & 2 & 3 & 8 \\ 9 & 5 & 1 & 3 & 6 & 4 & 2 & X & 8 & 7 \\ 5 & 1 & 6 & 2 & 4 & 7 & 3 & 8 & X & 9 \\ 1 & 2 & 3 & 4 & 5 & 6 & 8 & 7 & 9 & X \end{array}\right)$$

So it looks very plausible to me, that the completion can always be done for $n\geq 8$ and it is not unique for $n\geq 10$.

For the case $n=8$, with the precoloring you describe the completion you give is indeed unique. I checked by writing the corresponding boolean program and let a solver enumerate all solutions: there is only one.

For the case $n=10$, consider the pre-colored $K_{10}$ $$\left(\begin{array}{rrrrrrrrrr} X & & & & & 8 & 4 & 9 & 5 & 1 \\ & X & & & & & 9 & 5 & 1 & 2 \\ & & X & & & & & 1 & 6 & 3 \\ & & & X & & & & & 2 & 4 \\ & & & & X & & & & & 5 \\ 8 & & & & & X & & & & \\ 4 & 9 & & & & & X & & & \\ 9 & 5 & 1 & & & & & X & & \\ 5 & 1 & 6 & 2 & & & & & X & \\ 1 & 2 & 3 & 4 & 5 & & & & & X \end{array}\right)$$ This can be completed in $333$ ways, for example $$\left(\begin{array}{rrrrrrrrrr} X & 6 & 2 & 7 & 3 & 8 & 4 & 9 & 5 & 1 \\ 6 & X & 7 & 8 & 4 & 3 & 9 & 5 & 1 & 2 \\ 2 & 7 & X & 5 & 9 & 4 & 8 & 1 & 6 & 3 \\ 7 & 8 & 5 & X & 6 & 9 & 1 & 3 & 2 & 4 \\ 3 & 4 & 9 & 6 & X & 1 & 2 & 7 & 8 & 5 \\ 8 & 3 & 4 & 9 & 1 & X & 5 & 2 & 7 & 6 \\ 4 & 9 & 8 & 1 & 2 & 5 & X & 6 & 3 & 7 \\ 9 & 5 & 1 & 3 & 7 & 2 & 6 & X & 4 & 8 \\ 5 & 1 & 6 & 2 & 8 & 7 & 3 & 4 & X & 9 \\ 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & X \end{array}\right) $$ or $$ \left(\begin{array}{rrrrrrrrrr} X & 6 & 2 & 7 & 3 & 8 & 4 & 9 & 5 & 1 \\ 6 & X & 4 & 8 & 7 & 3 & 9 & 5 & 1 & 2 \\ 2 & 4 & X & 5 & 8 & 9 & 7 & 1 & 6 & 3 \\ 7 & 8 & 5 & X & 9 & 1 & 6 & 3 & 2 & 4 \\ 3 & 7 & 8 & 9 & X & 2 & 1 & 6 & 4 & 5 \\ 8 & 3 & 9 & 1 & 2 & X & 5 & 4 & 7 & 6 \\ 4 & 9 & 7 & 6 & 1 & 5 & X & 2 & 3 & 8 \\ 9 & 5 & 1 & 3 & 6 & 4 & 2 & X & 8 & 7 \\ 5 & 1 & 6 & 2 & 4 & 7 & 3 & 8 & X & 9 \\ 1 & 2 & 3 & 4 & 5 & 6 & 8 & 7 & 9 & X \end{array}\right)$$

So it looks very plausible to me, that the completion can always be done for $n\geq 8$ and it is not unique for $n\geq 10$.

I don't quite get what you mean by "distinct consecutive transversals", can you explain more clearly how the pre-coloring is done?

One question that I can answer: For the case $n=8$, with the precoloring you describe the completion you give is indeed unique. I checked by writing the corresponding boolean program and let a solver enumerate all solutions: there is only one.

I don't quite get what you mean by "distinct consecutive transversals", can you explain more clearly how the pre-coloring is done? I assume you mean fixing some values that don't contradict a valid coloring for the k outer diagonals.

For the case $n=8$, with the precoloring you describe the completion you give is indeed unique. I checked by writing the corresponding boolean program and let a solver enumerate all solutions: there is only one.

For the case $n=10$, consider the pre-colored $K_{10}$ $$\left(\begin{array}{rrrrrrrrrr} X & & & & & 8 & 4 & 9 & 5 & 1 \\ & X & & & & & 9 & 5 & 1 & 2 \\ & & X & & & & & 1 & 6 & 3 \\ & & & X & & & & & 2 & 4 \\ & & & & X & & & & & 5 \\ 8 & & & & & X & & & & \\ 4 & 9 & & & & & X & & & \\ 9 & 5 & 1 & & & & & X & & \\ 5 & 1 & 6 & 2 & & & & & X & \\ 1 & 2 & 3 & 4 & 5 & & & & & X \end{array}\right)$$ This can be completed in $333$ ways, for example $$\left(\begin{array}{rrrrrrrrrr} X & 6 & 2 & 7 & 3 & 8 & 4 & 9 & 5 & 1 \\ 6 & X & 7 & 8 & 4 & 3 & 9 & 5 & 1 & 2 \\ 2 & 7 & X & 5 & 9 & 4 & 8 & 1 & 6 & 3 \\ 7 & 8 & 5 & X & 6 & 9 & 1 & 3 & 2 & 4 \\ 3 & 4 & 9 & 6 & X & 1 & 2 & 7 & 8 & 5 \\ 8 & 3 & 4 & 9 & 1 & X & 5 & 2 & 7 & 6 \\ 4 & 9 & 8 & 1 & 2 & 5 & X & 6 & 3 & 7 \\ 9 & 5 & 1 & 3 & 7 & 2 & 6 & X & 4 & 8 \\ 5 & 1 & 6 & 2 & 8 & 7 & 3 & 4 & X & 9 \\ 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & X \end{array}\right) $$ or $$ \left(\begin{array}{rrrrrrrrrr} X & 6 & 2 & 7 & 3 & 8 & 4 & 9 & 5 & 1 \\ 6 & X & 4 & 8 & 7 & 3 & 9 & 5 & 1 & 2 \\ 2 & 4 & X & 5 & 8 & 9 & 7 & 1 & 6 & 3 \\ 7 & 8 & 5 & X & 9 & 1 & 6 & 3 & 2 & 4 \\ 3 & 7 & 8 & 9 & X & 2 & 1 & 6 & 4 & 5 \\ 8 & 3 & 9 & 1 & 2 & X & 5 & 4 & 7 & 6 \\ 4 & 9 & 7 & 6 & 1 & 5 & X & 2 & 3 & 8 \\ 9 & 5 & 1 & 3 & 6 & 4 & 2 & X & 8 & 7 \\ 5 & 1 & 6 & 2 & 4 & 7 & 3 & 8 & X & 9 \\ 1 & 2 & 3 & 4 & 5 & 6 & 8 & 7 & 9 & X \end{array}\right)$$

So it looks very plausible to me, that the completion can always be done for $n\geq 8$ and it is not unique for $n\geq 10$.