I would like to count the number of unique $k$-dimensional subspaces of the form $VD$, where $V\in \text{GL}(n,\mathbb Z/({p^m}))$ is some $n\times n$ matrix defined over the ring of integers modulo $p^m$, for some prime $p$ and integer $m$, i.e., $\mathbb Z/(p^m)$. Furthermore, $D$ is some predefined diagonal matrix with elements $d_i\in\{p^0,p^1, \dots,p^m\}$.
In other words, given $D=\text{diag}(\mathbf d)$, defining the set
$$S_{n,p^m}(\mathbf d)=\{VD : V\in \text{GL}(n,\mathbb Z/(p^m)),D=\text{diag}(\mathbf d) \},$$
what is the number of elements $S_{n,p^m}(\mathbf d)$ that are unique up to linear transformations on the right-hand side?
Defining $$\begin{aligned}T_{n,p^m}(\mathbf d)=S_{n,p^m}(\mathbf d)/GL(n,\mathbb Z/({p^m}))&=\{S_{n,p^m}(\mathbf d)R:R \in GL(n,\mathbb Z/(p^m))\}\\ &=\{\{XR:X\in S_{n,p^m}(\mathbf d)\}:R \in GL(n,\mathbb Z/(p^m))\},\end{aligned}$$ the question is to find the cardinality of $T_{n,p^m}(\mathbf d)$, i.e., $|T_{n,p^m}(\mathbf d)|$. The answer will be some function depending on $\mathbf d$, $n$ and $p^m$.
General Examples
- if $d_i=1$ for all $i$, then we have the trivial case of a single $n$-dimensional subspace.
- if $d_i=1$ for all $i\le k$ with $d_i=0$ otherwise, then we can count the number of $k$-dimensional subspaces on $\mathbb Z_{p^m}$ according to this answer.
For other choices of $D$ the problem becomes complicated.
Solution attempts
I originally tried to solve the problem by considering $V$ as defined in terms of $m$ matrices over the field $\mathbb Z_p$ but I am not sure this method avoids overcounting.
Alternatively, consider the column Hermite normal form of $VD$, which is a unique triangular matrix $H_{VD}=VDU$, with $U$ a unimodular matrix. If there exists some $R$ such that $VDR=V'D$ then the Hermite normal form of $VD$ and $V'D$ will be the same. Hence, we can count the number of Hermite normal form that exist.
The question then becomes, how many unique matrices $H$ exist, where each $H$ is a Hermite normal form of a matrix of the form $VD$?
Specific example
As a simple example of what I mean, consider $n=2$ with $p^m=2^4$. If I define $\mathbf d=(0,1)^T$, I want to count all the unique subspaces of the form $VD$ where $V=(\mathbf v_1,\mathbf v_2)$, where we see that $VD=(0,\mathbf v_2)$. Hence, in this case, the size of $T_{2,2^4}(\mathbf d)$, with $\mathbf d=(0,1)^T$ is equivalent to the number of subpaces of $GL(2,\mathbb Z/(16))$ consisting of only one vector $\mathbf v_2$. The question becomes, how many choices of $\mathbf v_2$ exist such that the set $\{(0,\mathbf v_2) R : R\in GL(n,\mathbb Z/({p^m}))\}$ is unique? In this case, this is equivalent to the question of the number of unique sets of the form $\{R^T\mathbf v_2 : GL(n,\mathbb Z/(p^m))\}$, which is in turn equivalent to asking the number of unique $1$-dimensional subspaces of the $n$-dimensional space $GL(n,\mathbb Z/({p^m}))$.
All the suborbits of $S_{n,p^m}({\bf d})$ under the right $GL$-action have equal sizes, since they are transitively permuted by the left $GL$-action, which conserves the sizes. (Since the orbits $O_U = UD \cdot GL$ and $O_V = VD \cdot GL$ are related by $O_U = U V^{-1} O_V$, we have $|O_U| \le |O_V|$).
The order of the group that acts is $|GL_n(\Bbb Z_{p^m})| = p^{(m-1) n^2} |GL_n(\Bbb Z_{p})| = p^{(m-1) n^2} \prod\limits_{i=0}^{n-1} (p^n - p^i)$, since a matrix over $\Bbb Z_{p^m}$ is invertible whenever its projection to $\Bbb Z_p$ is invertible (you can deduce this from the determinant invertibility criterion, or just construct the inverse explicitly in some other way than using the adjugate).
Let $D = I_{k_0} \oplus p I_{k_1} \oplus \ldots \oplus p^{m-1} I_{k_{m-1}} \oplus 0 I_{k_m}$ (matrix direct sum, $\sum k_i = n$).
Denote $R = \Bbb Z_{p^m}$.
(1) Over a commutative ring, a matrix is invertible whenever its determinant is invertible. The noninvertible elements of $R$ are exactly $p R$, so the determinant must project to a nonzero element of $\Bbb Z_p$.
(2) Define sets $S_U = U \cdot D \cdot GL$ for $U \in GL_n(R)$. Then left multiplication by another element $A$ of $GL$ permutes these sets: $A S_U = S_{AU}$, and we have $GL$ transitively left-acting on the set $\Sigma$ of these sets $S_U$. You asked for $T = |\Sigma|$, the length of the orbit, which is the index of the stabilizer of an arbitrary element of the orbit by the orbit-stabilizer theorem. So we can take $U$ the identity matrix and compute the left stabilizer of $D \cdot GL$.
Let's find the matrices $V$ such that $V \cdot D \cdot GL \subset D \cdot GL$. This is equivalent to existence of an invertible $K$ such that $V D = DK$.
Right multiplication by $D$ scales the columns, left multiplication scales rows. Divide our matrices into blocks so that the diagonal ones are square of sizes respectively $k_0, \ldots, k_m$.
The condition (on $Q$) "$\exists V, K$: $VD = DK = Q$" is equivalent for $Q$ to belong to the following set of block diagonal matrices, where by $p^k R$ I mean any submatrix with entries from that ideal (and the enumeration of block indices starts from $0$). Note that we don't yet assume $V, K$ to be invertible:
$$Q \in \begin{pmatrix} R & pR & p^2 R & \cdots & p^{m-1}R & 0\\ pR & pR & p^2 R & \cdots & p^{m-1}R & 0\\ p^2 R & p^2 R & p^2 R & \ldots & p^{m-1}R & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ p^{m-1}R & p^{m-1}R & p^{m-1}R & \cdots & p^{m-1}R & 0 \\ 0 & 0 & 0 & \cdots & 0 & 0 \end{pmatrix}$$
so, "dividing by the powers $p^k$", we can now treat the condition on $V$ "$\exists K$: $VD = DK$" ($V, K$ not necessarily invertible).
First note that in the leftmost (0-th) block column of $V$, which is scaled by $1$ when right-multiplied by $D$, the diagonal block $V_{10}$ cannot contain invertible elements (i.e. its entries are in $p R$) — otherwise $VD$ would also have an invertible element in the same block, and this would contradict the above analysis of possible $Q$'s. In the same way, the block $V_{20}$ has to consist only of elements that belong to $p^2 R$, and so on until we get that $V_{m0}$ should be zero.
The same procedure applies for the blocks that are below the diagonal in the next column: $D$ multiplies the block column by $p = p^1$, so $V_{21}$ has to consist of elements in $pR$ (for them to end up in $p^2 R$ in $VD$), and so on until $V_{m1}$ which is to be comprised of elements of $p^{m-1} R$. Carry out the same procedure for all block columns. Note that the elements of the diagonal or above-diagonal blocks of $V$ do not suffer from the conditions on $Q$, as they are being multiplied by a high power of $p$, so in $V$ we can put anything there. So the set of not necessarily invertible $V$'s is the following:
$$\begin{pmatrix} R & R & R & \cdots & R & R\\ pR & R & R & \cdots & R & R\\ p^2 R & p R & R & \ldots & R & R\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ p^{m-1}R & p^{m-2}R & p^{m-3}R & \cdots & R & R \\ 0 & p^{m-1}R & p^{m-2}R & \cdots & pR & R \end{pmatrix}$$
It is also straightforward to see the existence of $K$ for all matrices $V$ we found (again, both are not yet assumed to be invertible).
Now we have to ensure the condition $V \in GL$. The projection of any $V$ to $\Bbb Z_p$ is a block upper-triangular matrix; such matrices are invertible whenever each of their diagonal blocks is. Then the set of invertible $V$'s is:
$$\begin{pmatrix} GL & R & R & \cdots & R & R\\ pR & GL & R & \cdots & R & R\\ p^2 R & p R & GL & \ldots & R & R\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ p^{m-1}R & p^{m-2}R & p^{m-3}R & \cdots & GL & R \\ 0 & p^{m-1}R & p^{m-2}R & \cdots & pR & GL \end{pmatrix}$$
And, of course, for such an invertible $V$ there always exists an invertible $K$, if we take the first block row the same as in $VD$; in the second block row, the same as $VD$, but "divided" (arbitrarily) by $p$; and so on.
The total number of invertible $V$'s is thus
$$h = |R|^\alpha \times \prod\limits_{i = 0}^{m} |GL_{k_i}(R)| \times \prod\limits_{\ell = 1}^{m-1} p^{(m - \ell) \beta_\ell},$$
This is the order of the stabilizer, so $T_{n,p^m} = |GL_n(R)|/h$. As I wrote before, $|GL_a(R)| = p^{(m - 1)a^2} |GL_a(\Bbb Z_p)|$, so you can use this to expand the formulae.