I'll complete the argument that one cannot force a $n\times n$ determinant to be $1$ without fixing at least $\frac{n(n+1)}2$ of its entries, and with that number one must fix $n$ entries to $1$ and $\binom n2$ entries to $0$.
For the determinant to have a fixed value, all terms of the Leibniz formula must individually have a fixed value (since they all involve distinct monomials, and their sum is a constant polynomial). Clearly at least one fixed value must be $1$, arising for an even permutation $\sigma$. Then permuting the rows according to $\sigma$ we get another matrix with as many fixed entries, and the same value, and whose diagonal entries are all fixed to be $1$. Now for any $i\neq j$ neither of the entries $a_{i,j}$ and $a_{j,i}$ is fixed, we get a non-constant terms for $\sigma$ equal to the transposition of $i$ and $j$. So we must fix at least one of each of the $\binom n2$ such pairs of entries, proving we cannot succeed with less then $\frac{n(n+1)}2$ fixed entries. Moreover if we use exactly that number, then the entry fixed in each mentioned pair must be $0$ (or else the term for the transposition would still not be constant), so we have $n$ (diagonal) entries fixed to $1$ and $\binom n2$ (off-diagonal) entries fixed to $0$. Of course these entries could have been in different positions before permuting the rows.
Finally to show that we have essentially forced a triangular matrix, consider the oriented graph on the set $\{1,2,\ldots,n\}$, where there is an edge from $i$ to $j$ if $a_{i,j}$ is not fixed. Then this graph does not have oriented cycles, or else the term for that cyclic permutation would not be constant (the same argument we used for $2$-cycles). So the condition of having an oriented path from $i$ to $j$ defines a partial ordering of the points of the graph. This can be extended to a total ordering by adding directed edges, and it then has $\binom n2$ edges; but we supposed that there were $\binom n2$ unconstrained entries, so we had a total ordering from the outset. Now conjugating by the permutation that maps this total ordering to the natural ordering of $\{1,2,\ldots,n\}$ will make the matrix upper unitriangular.