Orbits of action of $GL(m, \mathbb{Z}_{p^n})$ on the set of all subgroups of ${\mathbb{Z}_{p^n}}^m$.

Question

Let $GL(m, \mathbb{Z}_{p^n})$ is the automorphism group of group $M = (\mathbb{Z}_{p^n})^m$. It is well known that every finite abelian group is(in a unique way) a direct product of cyclic groups of primary(the power of some prime number) order. For group G we will write $$type(G) = [{p_1}^{k_{11}}, \ldots, {p_1}^{k_{1l_1}}, \ldots, {p_s}^{k_{s1}}, \ldots, {p_s}^{k_{sl_s}}]$$ if $$G \cong C_{{p_1}^{k_{11}}} \oplus \ldots \oplus C_{{p_1}^{k_{1l_1}}} \oplus \ldots \oplus C_{{p_s}^{k_{s1}}} \oplus \ldots \oplus C_{{p_s}^{k_{sl_s}}}.$$ It is clear that for any subgroup $U < M$ and any automorphism $\varphi \in GL(m, \mathbb{Z}_{p^n})$ $$type(U) = type(\varphi(U)).$$ But does the set of all subgroups of the same type form the entire orbit of action of $GL(m, \mathbb{Z}_{p^n})$ on the set of all subgroups of $M$? For example, if n = 1 then we deal with the vector space $V_m = \mathbb{Z}_{p} \oplus \ldots \oplus \mathbb{Z}_{p}$. In this case, the answer is obviously yes.