Because of the requirement on the length, the configuration is fixed: $M$ lies on the ray $CB$ past $B$, and $N$ lies on the ray $ED$ past $D$. See the figure. Let $\alpha = \angle BAC$ and $\beta = \angle ABC$. Moreover, let $H$ be the orthocentre of the triangle (in other words: the intersection of $AD$, $BE$, and $CF$). Thales's theorem yields that $AFHE$, $BDHF$, $CEHD$, $ABDE$, $BCEF$, and $CAFD$ are cyclic. Because of the cyclic quadrilateral $ABDE$, we get $\angle CED = 180^\circ - \angle AED = \angle ABD = \beta$ and because of the cyclic quadrilateral $BCEF$, we get $\angle AEF = 180^\circ - \angle CEF = \angle CBF = \beta$. Analogously, $\angle CDE$ and $\angle BDF$ equal $\alpha$.
From $\angle CED = \beta = \angle AEF$ it follows that $\angle DEH = 90^\circ - \beta = \angle FEH$. Hence, $EH$ is the angular bisector of $\angle DEF$. Because $DE \parallel FM$, we get that $\angle MFE = 180^\circ - \angle FED = 180^\circ - 2(90^\circ - \beta) = 2\beta$. As $FN$ is the angular bisector of $\angle MFE$, we have $\angle EFN = \frac{1}{2} \cdot 2\beta = \beta$. Because $\angle FEH = 90^\circ - \beta$, we also see that $FN$ and $EH$ are perpendicular, hence $EH$ is not only the angular bisector in $\triangle FEN$, but it is also an altitude. Therefore, this line is also the perpendicular bisector of $FN$. As $B$ lies on this line, we get $|BF| = |BN|$.
We already saw that $\angle CDE = \alpha = \angle BDF$. Because $DE \parallel FM$, we also have $\angle BMF = \angle CDE = \alpha$, hence $\angle DMF = \angle BMF = \angle BDF = \angle MDF$. Thus, $|FM| = |FD|$.
Let $S$ be the intersection of $AC$ with $MF$. Then we have $\angle BFM = \angle AFS$ and because $DE \parallel FM$, we get $\angle CED = \angle CSF$. The exterior angle theorem in triangle $AFS$ yields that $\angle CSF = \angle SAF + \angle AFS = \alpha + \angle AFS$.

Combining everything, we obtain $\angle CED = \alpha + \angle BFM$. On the other hand, we knew that $\angle CED = \beta$, hence $\angle BFM = \beta - \alpha$. Moreover, we know that $\angle BMF = \alpha$. We conclude that $|BF| = |BM|$ if and only if $\beta - \alpha = \alpha$, or if and only if $\beta = 2\alpha$. Because we already know that $|BF| = |BN|$, we get: $B$ is the circumcentre of $\triangle FMN$ if and only if $\beta = 2\alpha$.
Before, we saw that $EH$ is the perpendicular bisector and altitude in triangle $EFN$, hence this triangle is isosceles with top angle $E$, which yields that $\angle DNF = \angle ENF = \angle EFN = \beta$. Moreover, we know that $\angle CDE = \alpha = \angle BDF$, from which it follows that $\angle NDF = \angle NDB + \angle BDF = \angle CDE + \angle BDF = 2\alpha$. Hence, $|FD| = |FN|$ if and only if $\beta = 2\alpha$. Because we already knew that $|FM| = |FD|$, we now get: $F$ is the circumcentre of $\triangle DMN$ if and only if $\beta = 2\alpha$.
We conclude that $F$ is the circumcentre of $\triangle DMN$ if and only if $B$ is the circumcentre of $\triangle FMN$, as both properties are equivalent to $\beta = 2\alpha$. $\square$