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I think it can be found the following way. If $R$ is a commutative local ring then every idempotent matrix is conjugate to a diagonal idempotent $\begin{pmatrix} I_r & 0\cr 0 &0\end{pmatrix}$ . The point is that projective modules are free for local rings. This lets you write $R^n$ as a direct sum of the image and the kernel, both of which are free.

So now one has to just compute the size of the stabilizer of the standard rank $r$ diagonal idempotent under the conjugation action of $\mathrm{GL}(R)$ for $R=\mathbb{Z}_{p^{k}}$.

Added. I believe the stabilizer of the rank $r$ idempotent for a local ring is $\mathrm{GL}(R,r)\times\mathrm{GL}(R,n-r)$ like in the field case and so a formula is easily found.

Added. I compute the answer for $n\times n$ matrices over $\mathbb{Z}_{p^k}$ to be $$\sum_{r=0}^n \frac{p^{2r(n-r)(k-1)}|\mathrm{GL}(n,p)|}{|\mathrm{GL}(r,p)|\mathrm{GL}(n-r,p)|}.$$ Here I use a matrix is invertible over a local ring iff it is over the residue field. The orders of the $\mathrm{GL}(m,p)$ are of course well known.

I think it can be found the following way. If $R$ is a commutative local ring then every idempotent matrix is conjugate to a diagonal idempotent $\begin{pmatrix} I_r & 0\cr 0 &0\end{pmatrix}$ . The point is that projective modules are free for local rings. This lets you write $R^n$ as a direct sum of the image and the kernel, both of which are free.

So now one has to just compute the size of the stabilizer of the standard rank $r$ diagonal idempotent under the conjugation action of $\mathrm{GL}(R)$ for $R=\mathbb{Z}_{p^{k}}$.

Added. I believe the stabilizer of the rank $r$ idempotent for a local ring is $\mathrm{GL}(R,r)\times\mathrm{GL}(R,n-r)$ like in the field case and so a formula is easily found.

Added. I compute the answer for $n\times n$ matrices over $\mathbb{Z}_{p^k}$ to be $$\sum_{r=0}^n \frac{p^{2r(n-r)(k-1)}|\mathrm{GL}(n,p)|}{|\mathrm{GL}(r,p)|\cdot|\mathrm{GL}(n-r,p)|}.$$ Here I use a matrix is invertible over a local ring iff it is over the residue field. The orders of the $\mathrm{GL}(m,p)$ are of course well known.

I think it can be found the following way. If R is a commutative local ring then every idempotent matrix is equivalent conjugate to a diagonal idempotent $\begin{pmatrix} I_r & 0\cr 0 &0\end{pmatrix}$ . The point is projective is free for local rings. This lets you write $R^n$ as a direct sum of the image and the kernel, both of which are free.

So now one has to just compute the size of the stabilizer of the standard rank r diagonal idempotent under the conjugation action of GL(R) for $R=Z_{p^{k}}$.

Added. I believe the stabilizer of the rank r idempotent for a local ring is GL(R,r)xGL(R,n-r)$ like in the field case and so a formula is easily found.

Added. I compute the answer for nxn matrices over $Z_{p^k}$ to be $$\sum_{r=0}^n \frac{p^{2r(n-r)(k-1)}|Gl(n,p)|}{|Gl(r,p)|Gl(n-r,p)|}.$$ Here I use a matrix is invertible over a local ring iff it is over the residue field. The orders of the Gl(m,p) are of course well known.

I think it can be found the following way. If $R$ is a commutative local ring then every idempotent matrix is conjugate to a diagonal idempotent $\begin{pmatrix} I_r & 0\cr 0 &0\end{pmatrix}$ . The point is that projective modules are free for local rings. This lets you write $R^n$ as a direct sum of the image and the kernel, both of which are free.

So now one has to just compute the size of the stabilizer of the standard rank $r$ diagonal idempotent under the conjugation action of $\mathrm{GL}(R)$ for $R=\mathbb{Z}_{p^{k}}$.

Added. I believe the stabilizer of the rank $r$ idempotent for a local ring is $\mathrm{GL}(R,r)\times\mathrm{GL}(R,n-r)$ like in the field case and so a formula is easily found.

Added. I compute the answer for $n\times n$ matrices over $\mathbb{Z}_{p^k}$ to be $$\sum_{r=0}^n \frac{p^{2r(n-r)(k-1)}|\mathrm{GL}(n,p)|}{|\mathrm{GL}(r,p)|\mathrm{GL}(n-r,p)|}.$$ Here I use a matrix is invertible over a local ring iff it is over the residue field. The orders of the $\mathrm{GL}(m,p)$ are of course well known.

I think it can be found the following way. If R is a commutative local ring then every idempotent matrix is equivalent to a diagonal idempotent $\begin{pmatrix} I_r & 0\cr 0 &0\end{pmatrix}$ . The point is projective is free for local rings. This lets you write $R^n$ as a direct sum of the image and the kernel, both of which are free.

So now one has to just compute the size of the stabilizer of the standard rank r diagonal idempotent under the conjugation action of GL(R) for $R=Z_{p^{k}}$.

Added. I believe the stabilizer of the rank r idempotent for a local ring is GL(R,r)xGL(R,n-r)$ like in the field case and so a formula is easily found.

Added. I compute the answer for nxn matrices over $Z_{p^k}$ to be $$\sum_{r=0}^n \frac{p^{2r(n-r)(k-1)}|Gl(n,p)|}{|Gl(r,p)|Gl(n-r,p)|}.$$ Here I use a matrix is invertible over a local ring iff it is over the residue field. The orders of the Gl(m,p) are of course well known.

I think it can be found the following way. If R is a commutative local ring then every idempotent matrix is equivalent conjugate to a diagonal idempotent $\begin{pmatrix} I_r & 0\cr 0 &0\end{pmatrix}$ . The point is projective is free for local rings. This lets you write $R^n$ as a direct sum of the image and the kernel, both of which are free.

So now one has to just compute the size of the stabilizer of the standard rank r diagonal idempotent under the conjugation action of GL(R) for $R=Z_{p^{k}}$.

Added. I believe the stabilizer of the rank r idempotent for a local ring is GL(R,r)xGL(R,n-r)$ like in the field case and so a formula is easily found.

Added. I compute the answer for nxn matrices over $Z_{p^k}$ to be $$\sum_{r=0}^n \frac{p^{2r(n-r)(k-1)}|Gl(n,p)|}{|Gl(r,p)|Gl(n-r,p)|}.$$ Here I use a matrix is invertible over a local ring iff it is over the residue field. The orders of the Gl(m,p) are of course well known.