I believe it is easier to first show when H exists for the second requirement, and then show if/when D exists for the first requirement given that H exists. All this depends upon different cases for the value of ${\sigma _2}$ and the relative sizes of m and n, so I parse the answer along those lines.
If ${\sigma _2} = 0$, then necessarily $H = \underline {\overline {\bf{0}} }$ for any values of m and n; then D exists only if ${\sigma _1} = 0$, and in that case, it can be any diagonal matrix.
If ${\sigma _2} \ne 0$ and $m < n$, then no H which satisfies the second requirement exists. The second requirement effectively requires H to be partial orthonormal, and no partial orthonormal matrix exists when rows have been deleted (i.e., $m < n$)
If ${\sigma _2} > 0$ and $m \ge n$, then H exists; any orthonormal matrix for which the requisite number of columns has been deleted and has then scaled by $\sqrt {{\sigma _2}} $ is a candidate. However, singular value decomposition then insures that the only possible value of D is $\left( {\frac{{{\sigma _1}}}{{{\sigma _2}}}} \right)I$. Any diagonal matrix other than this particular choice cannot result in ${\sigma _2}I$. In general, H will have no rows consisting of all zeros. In special cases, H may have up to $m - n$ rows consisting of all zeros; for any such row, the associated value in D may be arbitrarily chosen.
If ${\sigma _2} < 0$ and $m \ge n$, then H will exist but it will necessarily be complex. I believe only the same scaled D then exists (but I haven’t thought carefully about how all the different possible square roots of $ - I$ will play through the first requirement).
Note that in the first and third cases above, the arbitrary portions of D need not be diagonal.
