If you plot the periodogram in Mathematica, Nyquist will be in the middle of the frequencies.

x = Table[Sin[10π i/ 100],{i,100}];
ListPlot[x,Joined->True]
y = Abs[Fourier[x]];
ListPlot[y,Joined->True,PlotRange->All]

So the frequencies go from 0 up to Nyquist, which is in the middle.   Here, along the x-axis I'm just using the index number, not real frequencies.  If you don't know the sampling frequency, then you can just map the first n/2 points into "normalized" frequencies from 0 to 1, where 1 is by definition Nyquist.

n = 100
frequencies = Table[i/(n/2.),{i,1,n/2}]
data = Table[{frequencies[[i]],y[[i]]},{i,n/2}];
ListPlot[data,Joined->True,PlotRange->All]

yy = Table[0,{i,100}];
Do[yy[[i]] = y[[51-i]],{i,50,100}]
Do[yy[[i]] = y[[i+50]],{i,1,49}]
ListPlot[Abs[yy],Joined->True,PlotRange->All]
xax =  Table[2.i/n - 1.,{i,1,n}];
data = Table[{xax[[i]],yy[[i]]},{i,n}];
ListPlot[data,Joined->True,PlotRange->All]