UPDATE: There is an error at the end of this answer, addressed here and in comments. I am leaving the original answer here because it is discussed elsewhere.

Ugh, this turned out to be a nasty bit of brute forcing and I found myself grappling with notation before I got it (I think) right.

Write $J := (j_1,\dots,j_N)$ and $p^\otimes := p^{(1)} \otimes \dots \otimes p^{(N)}$ for the tensor product measure. Besides the obvious variations of the above notation, write $f^\otimes_{J^\land_m} := \prod_{\ell \ne m} f^{(\ell)}_{j_\ell}$. Finally, write $\partial_\ell \equiv \partial_{f_\ell}$.

Now for $\mbox{Var}_p \ne 0$ we have

$\partial_\ell \left( \mathcal{D}/\mbox{Var}_p \right) = 0 \iff -\partial_\ell \mathcal{D} \cdot \mbox{Var}_p + \mathcal{D} \cdot \partial_\ell \mbox{Var}_p = 0$

and

$\partial_\ell \mathcal{D} = -\sum_j \left( p_j Q_{j\ell} + p_\ell Q_{\ell j} \right) f_j,$

while if we assume w/l/o/g that $\sum_j p_j f_j = 0$ throughout then $\partial_\ell \mbox{Var}_p = 2p_\ell f_\ell$.

It follows that the equalities in the first equation above are equivalent to

$\sum_j (p_j Q_{j\ell} + p_\ell Q_{\ell j})f_j \cdot \mbox{Var}_p(f) + 2\mathcal{D}(f) \cdot p_\ell f_\ell = 0$.

For $F: [\dim Q^\otimes] \rightarrow \mathbb{R}$ (not necessarily of the form $f^\otimes \equiv f^{\otimes N}$), set $(F^\land_{L^\land_m})_j := F_{\ell_1,\dots,\ell_{m-1},j,\ell_{m+1},\dots,\ell_N}$. Now if $Q^{(m)} = c_m Q$ then

$\mathcal{D}^\otimes(F) = \frac{1}{2} \sum_{JK} p^\otimes_J Q^\otimes_{JK} (F_J - F_K)^2 = \frac{1}{2} \sum_{m,J^\land_m,j_m,k} p^\otimes_{J^\land_m} p_{j_m} c_m Q_{j_m k} \left ( (F^\land_{J^\land_m})_{j_m} - (F^\land_{J^\land_m})_{k} \right )^2$

$= \sum_m c_m \sum_{J^\land_m} p^\otimes_{J^\land_m} \mathcal{D}(F^\land_{J^\land_m})$

and

$\partial_L \mathcal{D}^\otimes(F) = -\sum_J (p^\otimes_J Q^\otimes_{JL} + p^\otimes_L Q^\otimes_{LJ}) F_J = -\sum_m c_m p^\otimes_{L^\land_m} \sum_j (p_j Q_{j \ell_m} + p_{\ell_m} Q_{\ell_m j})(F^\land_{L^\land_m})_j$.

Since $([f^\otimes]^\land_{L^\land_m})_j = f_j \cdot f^\otimes_{L^\land_m}$ and $\mathcal{D}([f^\otimes]^\land_{J^\land_m}) = (f^\otimes_{J^\land_m})^2 \cdot \mathcal{D}(f)$ it follows from the above that

$\left ( -\partial_L \mathcal{D}^\otimes \cdot \mbox{Var}_{p^\otimes} + \mathcal{D}^\otimes \cdot \partial_L \mbox{Var}_{p^\otimes} \right ) |_{f^\otimes}$

$= \sum_m c_m p^\otimes_{L^\land_m} \sum_j (p_j Q_{j \ell_m} + p_{\ell_m} Q_{\ell_m j})f_j \cdot f^\otimes_{L^\land_m} \cdot \mbox{Var}_p^N(f) + \sum_m c_m \sum_{J^\land_m} p^\otimes_{J^\land_m} (f^\otimes_{J^\land_m})^2 \cdot \mathcal{D}(f) \cdot 2 p^\otimes_L f^\otimes_L$

$= \mbox{Var}_{p}^{N-1}(f) \cdot \sum_m c_m p^\otimes_{L^\land_m} f^\otimes_{L^\land_m} \cdot \left [ \sum_j (p_j Q_{j \ell_m} + p_{\ell_m} Q_{\ell_m j})f_j \cdot \mbox{Var}_{p}(f) + 2\mathcal{D}(f) \cdot p_{\ell_m} f_{\ell_m} \right ]$.

Comparison with above shows that the term in brackets is zero iff $\partial_{\ell_m} \left( \mathcal{D}/\mbox{Var}_{p} \right) = 0$. Inspection of the above equations also shows that $\mathcal{D}^\otimes/\mbox{Var}_{p^\otimes}$ has a local minimum (resp. maximum) at $f^\otimes$ iff $\mathcal{D}/\mbox{Var}_p$ has a local minimum (resp. maximum) at $f$. Finally, the uniqueness of extrema follows from viewing $\mathcal{D}/\mbox{Var}_p$ as the (well-behaved) quotient of two homogeneous quadratic polynomials in the entries of $f$.

UPDATE: There is an error at the end of this answer, addressed here and in comments. I am leaving the original answer here because it is discussed elsewhere.

Ugh, this turned out to be a nasty bit of brute forcing and I found myself grappling with notation before I got it (I think) right.

Write $J := (j_1,\dots,j_N)$ and $p^\otimes := p^{(1)} \otimes \dots \otimes p^{(N)}$ for the tensor product measure. Besides the obvious variations of the above notation, write $f^\otimes_{J^\land_m} := \prod_{\ell \ne m} f^{(\ell)}_{j_\ell}$. Finally, write $\partial_\ell \equiv \partial_{f_\ell}$.

Now for $\mbox{Var}_p \ne 0$ we have

$\partial_\ell \left( \mathcal{D}/\mbox{Var}_p \right) = 0 \iff -\partial_\ell \mathcal{D} \cdot \mbox{Var}_p + \mathcal{D} \cdot \partial_\ell \mbox{Var}_p = 0$

and

$\partial_\ell \mathcal{D} = -\sum_j \left( p_j Q_{j\ell} + p_\ell Q_{\ell j} \right) f_j,$

while if we assume w/l/o/g that $\sum_j p_j f_j = 0$ throughout then $\partial_\ell \mbox{Var}_p = 2p_\ell f_\ell$.

It follows that the equalities in the first equation above are equivalent to

$\sum_j (p_j Q_{j\ell} + p_\ell Q_{\ell j})f_j \cdot \mbox{Var}_p(f) + 2\mathcal{D}(f) \cdot p_\ell f_\ell = 0$.

For $F: [\dim Q^\otimes] \rightarrow \mathbb{R}$ (not necessarily of the form $f^\otimes \equiv f^{\otimes N}$), set $(F^\land_{L^\land_m})_j := F_{\ell_1,\dots,\ell_{m-1},j,\ell_{m+1},\dots,\ell_N}$. Now if $Q^{(m)} = c_m Q$ then

$\mathcal{D}^\otimes(F) = \frac{1}{2} \sum_{JK} p^\otimes_J Q^\otimes_{JK} (F_J - F_K)^2 = \frac{1}{2} \sum_{m,J^\land_m,j_m,k} p^\otimes_{J^\land_m} p_{j_m} c_m Q_{j_m k} \left ( (F^\land_{J^\land_m})_{j_m} - (F^\land_{J^\land_m})_{k} \right )^2$

$= \sum_m c_m \sum_{J^\land_m} p^\otimes_{J^\land_m} \mathcal{D}(F^\land_{J^\land_m})$

and

$\partial_L \mathcal{D}^\otimes(F) = -\sum_J (p^\otimes_J Q^\otimes_{JL} + p^\otimes_L Q^\otimes_{LJ}) F_J = -\sum_m c_m p^\otimes_{L^\land_m} \sum_j (p_j Q_{j \ell_m} + p_{\ell_m} Q_{\ell_m j})(F^\land_{L^\land_m})_j$.

Since $([f^\otimes]^\land_{L^\land_m})_j = f_j \cdot f^\otimes_{L^\land_m}$ and $\mathcal{D}([f^\otimes]^\land_{J^\land_m}) = (f^\otimes_{J^\land_m})^2 \cdot \mathcal{D}(f)$ it follows from the above that

$\left ( -\partial_L \mathcal{D}^\otimes \cdot \mbox{Var}_{p^\otimes} + \mathcal{D}^\otimes \cdot \partial_L \mbox{Var}_{p^\otimes} \right ) |_{f^\otimes}$

$= \sum_m c_m p^\otimes_{L^\land_m} \sum_j (p_j Q_{j \ell_m} + p_{\ell_m} Q_{\ell_m j})f_j \cdot f^\otimes_{L^\land_m} \cdot \mbox{Var}_p^N(f) + \sum_m c_m \sum_{J^\land_m} p^\otimes_{J^\land_m} (f^\otimes_{J^\land_m})^2 \cdot \mathcal{D}(f) \cdot 2 p^\otimes_L f^\otimes_L$

$= \mbox{Var}_{p}^{N-1}(f) \cdot \sum_m c_m p^\otimes_{L^\land_m} f^\otimes_{L^\land_m} \cdot \left [ \sum_j (p_j Q_{j \ell_m} + p_{\ell_m} Q_{\ell_m j})f_j \cdot \mbox{Var}_{p}(f) + 2\mathcal{D}(f) \cdot p_{\ell_m} f_{\ell_m} \right ]$.

Comparison with above shows that the term in brackets is zero iff $\partial_{\ell_m} \left( \mathcal{D}/\mbox{Var}_{p} \right) = 0$. Inspection of the above equations also shows that $\mathcal{D}^\otimes/\mbox{Var}_{p^\otimes}$ has a local minimum (resp. maximum) at $f^\otimes$ iff $\mathcal{D}/\mbox{Var}_p$ has a local minimum (resp. maximum) at $f$. Finally, the uniqueness of extrema follows from viewing $\mathcal{D}/\mbox{Var}_p$ as the (well-behaved) quotient of two homogeneous quadratic polynomials in the entries of $f$.

Ugh, this turned out to be a nasty bit of brute forcing and I found myself grappling with notation before I got it (I think) right.

Write $J := (j_1,\dots,j_N)$ and $p^\otimes := p^{(1)} \otimes \dots \otimes p^{(N)}$ for the tensor product measure. Besides the obvious variations of the above notation, write $f^\otimes_{J^\land_m} := \prod_{\ell \ne m} f^{(\ell)}_{j_\ell}$. Finally, write $\partial_\ell \equiv \partial_{f_\ell}$.

Now for $\mbox{Var}_p \ne 0$ we have

$\partial_\ell \left( \mathcal{D}/\mbox{Var}_p \right) = 0 \iff -\partial_\ell \mathcal{D} \cdot \mbox{Var}_p + \mathcal{D} \cdot \partial_\ell \mbox{Var}_p = 0$

and

$\partial_\ell \mathcal{D} = -\sum_j \left( p_j Q_{j\ell} + p_\ell Q_{\ell j} \right) f_j,$

while if we assume w/l/o/g that $\sum_j p_j f_j = 0$ throughout then $\partial_\ell \mbox{Var}_p = 2p_\ell f_\ell$.

It follows that the equalities in the first equation above are equivalent to

$\sum_j (p_j Q_{j\ell} + p_\ell Q_{\ell j})f_j \cdot \mbox{Var}_p(f) + 2\mathcal{D}(f) \cdot p_\ell f_\ell = 0$.

For $F: [\dim Q^\otimes] \rightarrow \mathbb{R}$ (not necessarily of the form $f^\otimes \equiv f^{\otimes N}$), set $(F^\land_{L^\land_m})_j := F_{\ell_1,\dots,\ell_{m-1},j,\ell_{m+1},\dots,\ell_N}$. Now if $Q^{(m)} = c_m Q$ then

$\mathcal{D}^\otimes(F) = \frac{1}{2} \sum_{JK} p^\otimes_J Q^\otimes_{JK} (F_J - F_K)^2 = \frac{1}{2} \sum_{m,J^\land_m,j_m,k} p^\otimes_{J^\land_m} p_{j_m} c_m Q_{j_m k} \left ( (F^\land_{J^\land_m})_{j_m} - (F^\land_{J^\land_m})_{k} \right )^2$

$= \sum_m c_m \sum_{J^\land_m} p^\otimes_{J^\land_m} \mathcal{D}(F^\land_{J^\land_m})$

and

$\partial_L \mathcal{D}^\otimes(F) = -\sum_J (p^\otimes_J Q^\otimes_{JL} + p^\otimes_L Q^\otimes_{LJ}) F_J = -\sum_m c_m p^\otimes_{L^\land_m} \sum_j (p_j Q_{j \ell_m} + p_{\ell_m} Q_{\ell_m j})(F^\land_{L^\land_m})_j$.

Since $([f^\otimes]^\land_{L^\land_m})_j = f_j \cdot f^\otimes_{L^\land_m}$ and $\mathcal{D}([f^\otimes]^\land_{J^\land_m}) = (f^\otimes_{J^\land_m})^2 \cdot \mathcal{D}(f)$ it follows from the above that

$\left ( -\partial_L \mathcal{D}^\otimes \cdot \mbox{Var}_{p^\otimes} + \mathcal{D}^\otimes \cdot \partial_L \mbox{Var}_{p^\otimes} \right ) |_{f^\otimes}$

$= \sum_m c_m p^\otimes_{L^\land_m} \sum_j (p_j Q_{j \ell_m} + p_{\ell_m} Q_{\ell_m j})f_j \cdot f^\otimes_{L^\land_m} \cdot \mbox{Var}_p^N(f) + \sum_m c_m \sum_{J^\land_m} p^\otimes_{J^\land_m} (f^\otimes_{J^\land_m})^2 \cdot \mathcal{D}(f) \cdot 2 p^\otimes_L f^\otimes_L$

$= \mbox{Var}_{p}^{N-1}(f) \cdot \sum_m c_m p^\otimes_{L^\land_m} f^\otimes_{L^\land_m} \cdot \left [ \sum_j (p_j Q_{j \ell_m} + p_{\ell_m} Q_{\ell_m j})f_j \cdot \mbox{Var}_{p}(f) + 2\mathcal{D}(f) \cdot p_{\ell_m} f_{\ell_m} \right ]$.

Comparison with above shows that the term in brackets is zero iff $\partial_{\ell_m} \left( \mathcal{D}/\mbox{Var}_{p} \right) = 0$. Inspection of the above equations also shows that $\mathcal{D}^\otimes/\mbox{Var}_{p^\otimes}$ has a local minimum (resp. maximum) at $f^\otimes$ iff $\mathcal{D}/\mbox{Var}_p$ has a local minimum (resp. maximum) at $f$. Finally, the uniqueness of extrema follows from viewing $\mathcal{D}/\mbox{Var}_p$ as the (well-behaved) quotient of two homogeneous quadratic polynomials in the entries of $f$.

UPDATE: There is an error at the end of this answer, addressed here and in comments. I am leaving the original answer here because it is discussed elsewhere.

Ugh, this turned out to be a nasty bit of brute forcing and I found myself grappling with notation before I got it (I think) right.

Write $J := (j_1,\dots,j_N)$ and $p^\otimes := p^{(1)} \otimes \dots \otimes p^{(N)}$ for the tensor product measure. Besides the obvious variations of the above notation, write $f^\otimes_{J^\land_m} := \prod_{\ell \ne m} f^{(\ell)}_{j_\ell}$. Finally, write $\partial_\ell \equiv \partial_{f_\ell}$.

Now for $\mbox{Var}_p \ne 0$ we have

$\partial_\ell \left( \mathcal{D}/\mbox{Var}_p \right) = 0 \iff -\partial_\ell \mathcal{D} \cdot \mbox{Var}_p + \mathcal{D} \cdot \partial_\ell \mbox{Var}_p = 0$

and

$\partial_\ell \mathcal{D} = -\sum_j \left( p_j Q_{j\ell} + p_\ell Q_{\ell j} \right) f_j,$

while if we assume w/l/o/g that $\sum_j p_j f_j = 0$ throughout then $\partial_\ell \mbox{Var}_p = 2p_\ell f_\ell$.

It follows that the equalities in the first equation above are equivalent to

$\sum_j (p_j Q_{j\ell} + p_\ell Q_{\ell j})f_j \cdot \mbox{Var}_p(f) + 2\mathcal{D}(f) \cdot p_\ell f_\ell = 0$.

For $F: [\dim Q^\otimes] \rightarrow \mathbb{R}$ (not necessarily of the form $f^\otimes \equiv f^{\otimes N}$), set $(F^\land_{L^\land_m})_j := F_{\ell_1,\dots,\ell_{m-1},j,\ell_{m+1},\dots,\ell_N}$. Now if $Q^{(m)} = c_m Q$ then

$\mathcal{D}^\otimes(F) = \frac{1}{2} \sum_{JK} p^\otimes_J Q^\otimes_{JK} (F_J - F_K)^2 = \frac{1}{2} \sum_{m,J^\land_m,j_m,k} p^\otimes_{J^\land_m} p_{j_m} c_m Q_{j_m k} \left ( (F^\land_{J^\land_m})_{j_m} - (F^\land_{J^\land_m})_{k} \right )^2$

$= \sum_m c_m \sum_{J^\land_m} p^\otimes_{J^\land_m} \mathcal{D}(F^\land_{J^\land_m})$

and

$\partial_L \mathcal{D}^\otimes(F) = -\sum_J (p^\otimes_J Q^\otimes_{JL} + p^\otimes_L Q^\otimes_{LJ}) F_J = -\sum_m c_m p^\otimes_{L^\land_m} \sum_j (p_j Q_{j \ell_m} + p_{\ell_m} Q_{\ell_m j})(F^\land_{L^\land_m})_j$.

Since $([f^\otimes]^\land_{L^\land_m})_j = f_j \cdot f^\otimes_{L^\land_m}$ and $\mathcal{D}([f^\otimes]^\land_{J^\land_m}) = (f^\otimes_{J^\land_m})^2 \cdot \mathcal{D}(f)$ it follows from the above that

$\left ( -\partial_L \mathcal{D}^\otimes \cdot \mbox{Var}_{p^\otimes} + \mathcal{D}^\otimes \cdot \partial_L \mbox{Var}_{p^\otimes} \right ) |_{f^\otimes}$

$= \sum_m c_m p^\otimes_{L^\land_m} \sum_j (p_j Q_{j \ell_m} + p_{\ell_m} Q_{\ell_m j})f_j \cdot f^\otimes_{L^\land_m} \cdot \mbox{Var}_p^N(f) + \sum_m c_m \sum_{J^\land_m} p^\otimes_{J^\land_m} (f^\otimes_{J^\land_m})^2 \cdot \mathcal{D}(f) \cdot 2 p^\otimes_L f^\otimes_L$

$= \mbox{Var}_{p}^{N-1}(f) \cdot \sum_m c_m p^\otimes_{L^\land_m} f^\otimes_{L^\land_m} \cdot \left [ \sum_j (p_j Q_{j \ell_m} + p_{\ell_m} Q_{\ell_m j})f_j \cdot \mbox{Var}_{p}(f) + 2\mathcal{D}(f) \cdot p_{\ell_m} f_{\ell_m} \right ]$.

Comparison with above shows that the term in brackets is zero iff $\partial_{\ell_m} \left( \mathcal{D}/\mbox{Var}_{p} \right) = 0$. Inspection of the above equations also shows that $\mathcal{D}^\otimes/\mbox{Var}_{p^\otimes}$ has a local minimum (resp. maximum) at $f^\otimes$ iff $\mathcal{D}/\mbox{Var}_p$ has a local minimum (resp. maximum) at $f$. Finally, the uniqueness of extrema follows from viewing $\mathcal{D}/\mbox{Var}_p$ as the (well-behaved) quotient of two homogeneous quadratic polynomials in the entries of $f$.