As [Henrik Schumacher][1] correctly [points out][2] in the comments, the stars on the plot produced by the [naive application][3] of the undocumented `MeshFunctions -> {"ArcLength"}` option 

>  do not look equidistantly spaced with respect to arclength

It took me a while to figure out what happens. The reason is that scales in the vertical and horizontal directions of the plot differ by an order of magnitude. I tried other plotting functions like `ListPlot`, `Plot`, `ParametricPlot` and found that all of them also suffer from this issue.

The following approach solves the problem *almost exactly*. Small inaccuracy comes from the fact that I use `AspectRatio` of the whole graphics instead of the actual aspect ratio of the plotting range with `PlotRangePadding` included. But visually it is hardly noticeable. From the other hand, it isn't too difficult to improve this approach for obtaining exact result.

We start by defining our `data1`:

    b1={1.8743,1.8784,1.88248,1.89049,1.89828,1.90587,1.91327,1.96335,2.03035,2.12536,2.23701,2.30098,2.34255,2.37175,2.3934,2.42334,2.44307,2.48725,2.51208,2.5173,2.52799,2.53164,2.53348,2.53533,2.53625,2.53745,2.53793,2.53894,2.53909,2.53543};
    ks1={0.01,1.,2.,4.,6.,8.,10.,25.,50.,100.,200.,300.,400.,500.,600.,800.,1000.,2000.,4000.,5000.,10000.,15000.,20000.,30000.,40000.,70000.,100000.,1.*10^6,1.*10^8,1.*10^12};
    data1 = Transpose[{ks1, b1}];

Now define future `AspectRatio` of the plot and log-transformed `data1` (the `Graphics` produced by `ListLogLinearPlot` contains log-transformed data: it is just `Ticks` what creates the illusion of logarithmic plot):

    aspectRatio = 1./GoldenRatio;
    data1Log = {Log@#1, #2} & @@@ data1;

Now create rescaling function and its inverse:

    t = RescalingTransform[
       MinMax /@ Transpose[data1Log], {{0, 1}, {0, aspectRatio}}];
    tInv = InverseFunction[t];

This function makes vertical and horizontal scales *almost* equal (if necessary one can improve it for obtaining exact result).

Then we use `MeshFunctions -> {"ArcLength"}` for generating the equidistantly spaced points with respect to arclength:

    s2 = ListLinePlot[t[data1Log], PlotRange -> All, 
       MeshFunctions -> {"ArcLength"}, Mesh -> True];

Extracting the equidistant points and performing the inverse coordinate transform:

    pts = tInv @@@ Cases[Normal[s2], _Point, -1];

Now we can use these coordinates for placing the stars in `Epilog`:

    ListLogLinearPlot[data1, Joined -> True, 
     PlotStyle -> {Black, Thickness[0.01]}, AxesStyle -> Black, PlotRange -> All,  
     Epilog -> {Red, Translate[Inset[Style["☆", 15, Bold], {0, 0}], pts]},
     AspectRatio -> aspectRatio]

> [![plot][4]][4]

Voila! :)

----------

**UPDATE**: I just accidentally found that I [knew][5] about this feature of `MeshFunctions -> {"ArcLength"}` already several years ago but completely forgot due to non-use of this functionality. I can add to that old post that now we have the wonderful [`GraphicsInformation`][6] function by [Carl Woll][7] which simplifies things considerably.


  [1]: https://mathematica.stackexchange.com/users/38178/henrik-schumacher
  [2]: https://mathematica.stackexchange.com/questions/188874/tracing-the-existing-curve-with-star-symbol#comment492389_188875
  [3]: https://mathematica.stackexchange.com/a/188875/280
  [4]: https://i.sstatic.net/irElC.png
  [5]: https://mathematica.stackexchange.com/a/64345/280
  [6]: https://mathematica.stackexchange.com/a/138907/280
  [7]: https://mathematica.stackexchange.com/users/45431/carl-woll