The answer is $\boxed5$. Proof: Since the sines of all the angles are equal, this means all the angles must be either $x^\circ$ or $(180-x)^\circ$, for some constant $x$. First, we show that $n=3$ or $n=4$ is not possible. If $n=3$, then the angles all must be $x^\circ,x^\circ,x^\circ$. This gives $$x+x+x=180\implies x=60$$This shows the triangle is equilateral, so its impossible for it to have different side lengths. If $n=4$, then the angles all must be either $x^\circ,x^\circ,x^\circ,x^\circ$ or $x^\circ,x^\circ,(180-x)^\circ,x^\circ$. This shows that the quadrilaterl is either a parallelogram or an isosceles trapezoid, so its also impossible for it to have different side lengths. Next, we show that $5$ is possible. Consider the pentagon with side lengths $1,2,4,3,6$ respectively, with angles $120^\circ,120^\circ,120^\circ,120^\circ,60^\circ$. [asy][asy]size(200);pair A=(0,0),B=A+dir(120),C=B+2*dir(60),D=C+4*dir(0),E=D+3*dir(-60); draw(A--B--C--D--E--A); label("$1$",(A+B)/2,dir(210));label("$2$",(B+C)/2,dir(150));label("$4$",(C+D)/2,dir(90));label("$3$",(D+E)/2,dir(30));label("$6$",(E+A)/2,dir(-90));[/asy][/asy]

Take points $A,B,C,D$ as such: Points $A$ and $B$ are such that $AB=6$. Now points $C$ and $D$ are such that $\angle CAB=120$, $AC=1$ and $\angle DBA=60$, $DB=3$. Now construct point $F$ on $DB$ such that $FC\parallel AB$. Now $ABFC$ is a parallelogram and so $FB=CA=1$ and $CF=AB=6$. Construct point E such that $CFDE$ is an isosceles trapezoid, so $CE=2$. We just have to prove that $ED=4$ to have completed the proof. Take points $X$ and $Y$ to be the feet of the perpendiculars from points $E$ and $D$ to $CF$ respectively. Now in $\triangle CEX: \angle CEX=30, \angle ECX=60$ , so $XC=\frac{1}{2}CE=1$. Symmetrically $YF=1$. Now $XYDE$ is a rectangle and so $ED=XY=CF-XC-YF=6-1-1=4$ and we are done.