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I will sharpen my earlier comment a little. The number of such matrices is divisible by $(p-1)^{2n-2}$ (though it may be zero). Let $T$ be the group of invertible diagonal matrices, and let $\mathcal{M}$ denote the set of invertible matrices of full rank ( though not necessarily determinant 1). Then $M \to TMU^{-1}$ defines an action of $T \times T$ on $\mathcal{M},$ and the only pairs $(T,U)$ which fix any element are those with $T = U$ both scalar. Hence $(T \times T)/{\rm diag} (S)$ acts semi-regularly on $\mathcal{M},$ where $S$ is the subgroup of $T$ consisting of scalars. Thus $|\mathcal{M}|$ is divisible by $(p-1)^{2n-1}$, and hence, as user74230 notes in comments, the number of matrices in $\mathcal{M}$ which have determinant $1$ is a multiple of $(p-1)^{2n-2}.$ Also, the number of such matrices is divisible by $\frac{n!}{2},$ since $A_{n}$ acts by right multiplication (in its representation by permutation matrices) on such matrices, and no non-identity element has any fixed point.

Actually, I realise that this implies that when $n \geq p,$ the number of such matrices is divisible by $p.$ For when $p =2$ the number is zero, and when $p$ is odd and $n \geq p,$ then $|A_{n}|$ is divisible by $p^{\frac{n - \sigma_{p}(n)}{p-1}}$, where $\sigma_{p}(n)$ is the sum of the digit in the base $p$ expansion of $n.$

I will sharpen my earlier comment a little. The number of such matrices is divisible by $(p-1)^{2n-2}$ (though it may be zero). Let $T$ be the group of invertible diagonal matrices, and let $\mathcal{M}$ denote the set of invertible matrices of full rank ( though not necessarily determinant 1). Then $M \to TMU^{-1}$ defines an action of $T \times T$ on $\mathcal{M},$ and the only pairs $(T,U)$ which fix any element are those with $T = U$ both scalar. Hence $(T \times T)/{\rm diag} (S)$ acts semi-regularly on $\mathcal{M},$ where $S$ is the subgroup of $T$ consisting of scalars. Thus $|\mathcal{M}|$ is divisible by $(p-1)^{2n-1}$, and hence, as user74230 notes in comments, the number of matrices in $\mathcal{M}$ which have determinant $1$ is a multiple of $(p-1)^{2n-2}.$ Also, the number of such matrices is divisible by $\frac{n!}{2},$ since $A_{n}$ acts by right multiplication (in its representation by permutation matrices) on such matrices, and no non-identity element has any fixed point.

Actually, I realise that this implies that when $n \geq p,$ the number of such matrices is divisible by $p.$ For when $p =2$ the number is zero, and when $p$ is odd and $n \geq p,$ then $|A_{n}|$ is divisible by $p^{\frac{n - \sigma_{p}(n)}{p-1}}$, where $\sigma_{p}(n)$ is the sum of the digits in the base $p$ expansion of $n.$

I will sharpen my earlier comment a little. The number of such matrices is divisible by $(p-1)^{2n-2}$ (though it may be zero). Let $T$ be the group of invertible diagonal matrices, and let $\mathcal{M}$ denote the set of invertible matrices of full rank ( though not necessarily determinant 1). Then $M \to TMU^{-1}$ defines an action of $T \times T$ on $\mathcal{M},$ and the only pairs $(T,U)$ which fix any element are those with $T = U$ both scalar. Hence $(T \times T)/{\rm diag} (S)$ acts semi-regularly on $\mathcal{M},$ where $S$ is the subgroup of $T$ consisting of scalars. Thus $|\mathcal{M}|$ is divisible by $(p-1)^{2n-1}$, and hence, as user74230 notes in comments, the number of matrices in $\mathcal{M}$ which have determinant $1$ is a multiple of $(p-1)^{2n-2}.$ Also, the number of such matrices is divisible by $\frac{n!}{2},$ since $A_{n}$ acts by right multiplication (in its representation by permutation matrices) on such matrices, and no non-identity element has any fixed point.

Actually, I realise that this implies that when $n \geq p,$ the number of such matrices is divisible by $p.$ For when $p =2$ the number is zero, and when $p$ is odd and $n \geq p,$ then $|A_{n}|$ is divisible by $p.$

I will sharpen my earlier comment a little. The number of such matrices is divisible by $(p-1)^{2n-2}$ (though it may be zero). Let $T$ be the group of invertible diagonal matrices, and let $\mathcal{M}$ denote the set of invertible matrices of full rank ( though not necessarily determinant 1). Then $M \to TMU^{-1}$ defines an action of $T \times T$ on $\mathcal{M},$ and the only pairs $(T,U)$ which fix any element are those with $T = U$ both scalar. Hence $(T \times T)/{\rm diag} (S)$ acts semi-regularly on $\mathcal{M},$ where $S$ is the subgroup of $T$ consisting of scalars. Thus $|\mathcal{M}|$ is divisible by $(p-1)^{2n-1}$, and hence, as user74230 notes in comments, the number of matrices in $\mathcal{M}$ which have determinant $1$ is a multiple of $(p-1)^{2n-2}.$ Also, the number of such matrices is divisible by $\frac{n!}{2},$ since $A_{n}$ acts by right multiplication (in its representation by permutation matrices) on such matrices, and no non-identity element has any fixed point.

Actually, I realise that this implies that when $n \geq p,$ the number of such matrices is divisible by $p.$ For when $p =2$ the number is zero, and when $p$ is odd and $n \geq p,$ then $|A_{n}|$ is divisible by $p^{\frac{n - \sigma_{p}(n)}{p-1}}$, where $\sigma_{p}(n)$ is the sum of the digit in the base $p$ expansion of $n.$

I will sharpen my earlier comment a little. The number of such matrices is divisible by $(p-1)^{2n-2}$ (though it may be zero). Let $T$ be the group of invertible diagonal matrices, and let $\mathcal{M}$ denote the set of invertible matrices of full rank ( though not necessarily determinant 1). Then $M \to TMU^{-1}$ defines an action of $T \times T$ on $\mathcal{M},$ and the only pairs $(T,U)$ which fix any element are those with $T = U$ both scalar. Hence $(T \times T)/{\rm diag} (S)$ acts semi-regularly on $\mathcal{M},$ where $S$ is the subgroup of $T$ consisting of scalars. Thus $|\mathcal{M}|$ is divisible by $(p-1)^{2n-1}$, and hence, as user74230 notes in comments, the number of matrices in $\mathcal{M}$ which have determinant $1$ is a multiple of $(p-1)^{2n-2}.$ Also, the number of such matrices is divisible by $\frac{n!}{2},$ since $A_{n}$ acts by right multiplication such matrices and no non-identity element has any fixed point.

I also remark that the general problem looks quite messy to me for $n >2.$ Having chosen the first $k$ linearly independent rows with all non-zero entries, we have $(p-1)^{n} - f(k)$ ways to complete the $k+1$-st row, where $f(k)$ is the number of row vectors in the span of the first $k$-rows which have all non-zero entries. But it is not obvious (and is probably not true) that the number $f(k)$ is independent of the choice of the first $k$-rows. Also, the last row chosen may be multiplied by any non-zero scalar, and one and only one choice of scalar makes the whole matrix have determinant $1.$

I will sharpen my earlier comment a little. The number of such matrices is divisible by $(p-1)^{2n-2}$ (though it may be zero). Let $T$ be the group of invertible diagonal matrices, and let $\mathcal{M}$ denote the set of invertible matrices of full rank ( though not necessarily determinant 1). Then $M \to TMU^{-1}$ defines an action of $T \times T$ on $\mathcal{M},$ and the only pairs $(T,U)$ which fix any element are those with $T = U$ both scalar. Hence $(T \times T)/{\rm diag} (S)$ acts semi-regularly on $\mathcal{M},$ where $S$ is the subgroup of $T$ consisting of scalars. Thus $|\mathcal{M}|$ is divisible by $(p-1)^{2n-1}$, and hence, as user74230 notes in comments, the number of matrices in $\mathcal{M}$ which have determinant $1$ is a multiple of $(p-1)^{2n-2}.$ Also, the number of such matrices is divisible by $\frac{n!}{2},$ since $A_{n}$ acts by right multiplication (in its representation by permutation matrices) on such matrices, and no non-identity element has any fixed point.

Actually, I realise that this implies that when $n \geq p,$ the number of such matrices is divisible by $p.$ For when $p =2$ the number is zero, and when $p$ is odd and $n \geq p,$ then $|A_{n}|$ is divisible by $p.$