Solution:

Answer: $2\,1/3$

Let the square be $ABCD$. Let the vertices of the rhombus be $P$ on $AB$, $Q$ on $AD$, and $R$ on $BC$. We require the locus of the fourth vertex $S$ of the rhombus. Suppose $P$ is a distance $x$ from $B$. We may take $x \leq 1/2$, since the locus for $x > 1/2$ is just the reflection of the locus for $x < 1/2$. Then since $PR$ is parallel to $QS$, $S$ is a distance $x$ from the line $AD$. Also, by continuity, as $Q$ varies over $AD$ (with $P$ fixed a distance $x$ from $B$), the locus of $S$ is a line segment.

The two extreme positions for $S$ occur when $Q$ coincides with $A$ and when $R$ coincides with $C$. When $Q$ coincides with $A$ the rhombus has side $1 - x$. Hence $BR^2 = (1 - x)^2 - x^2 = 1 - 2x$. In this case $SR$ is parallel to $AB$, so the distance of $S$ from $AB$ is $\sqrt{1 - 2x}$. When $R$ coincides with $C$, the rhombus has side $\sqrt{1 + x^2}$, so $AQ^2 = 1 + x^2 - (1 - x)^2 = 2x$. Hence the distance of $S$ from $AB$ is $1 + \sqrt{2x}$.

Thus the locus of $S$ over all possible rhombi is the interior of a curvilinear quadrilateral with vertices $MDNC$, where $M$ is the midpoint of $AB$ and $N$ is the reflection of $M$ in $CD$. Moreover the curve from $M$ to $C$ is just the translate of the curve from $D$ to $N$, for if we put $y = 1/2 - x$, then $\sqrt{1 - 2x}$ becomes $\sqrt{2y}$. Thus if $L$ is the midpoint of $CD$, then the area in the $MLC$ plus the area in $DLN$ is just $1/2$, and the total area of the curvilinear quadrilateral is $1$.

However, the arrangement of the vertices discussed above is not the only one. The order of vertices above is $PQSR$. We could also have $PQRS$ or $PSQR$. In either case $QR$ is a side rather than a diagonal of the rhombus. We consider the case $PQRS$ (the case $PSQR$ is just the reflection in the line $MN$). As before it is convenient to keep $P$ fixed, but this time we take $x$ to be the distance $AP$. Take $y$ to be the distance $AQ$.

As before we find that $S$ must lie on a line parallel to $BC$ a distance $x$ from it (on the other side to $AD$). Again we find that for fixed $P$, the locus of $S$ is a segment of this line. If we assume that $AQ > BR$, then the two extreme positions are (1) $QR$ parallel to $AB$, giving $S$ on the line $AB$, (2) $Q$ at $D$, giving $S$ a distance $x$ from the line $AB$. So as $x$ varies from $0$ to $1$ we get a right-angled triangle sides $1$, $1$ and $\sqrt{2}$ and area $1/2$. However, we can also have $BR > AQ$. This gives points below the line $AB$. The extreme position is with $R$ at $C$. Suppose $QD = y$. Then $1 + y^2 = x^2 + (1 - y)^2$, so $y = x^2/2$. This gives $S$ a distance $y$ below the line $AB$. This gives an additional area of $1/6$ (by calculus - integrate $x^2/2$ from $0$ to $1$; I do not see how to do it without).

The triangle and the curvilinear triangle together form a curvilinear triangle area $1/2 + 1/6 = 2/3$. There is an identical triangle formed by reflection in $MN$. Thus the total area is $1 + 2/3 + 2/3 = 2\,1/3$.