Owen and Cecil are right: the source (transmitter) has no effect whatever
on the VSWR on the line.

That isn't just an assertion - it is part of the bedrock transmission line
theory. Owen referred to "reputable textbooks", one of which would surely
be 'Theory and Problems of Transmission Lines' by R A Chipman [1]. This
book gains a lot of its reputation from its very complete mathematical
development of the theory, showing all the detailed working.

I am sorry but you are not correct, I have not read Chipman so I cannot
comment on his analysis or your interpretation of his results, but my
understanding , practical experiments and CAD analysis would lead me to
disagree.

If we take the situation where the source is matched (50ohms) to the 5.35
wavelength transmission line (lossless to simplify things) with a 100ohm
load, I agree that the vswr is 4:1, unchanging with frequency.

Plotted on a Smith Chart when swept against frequency this gives a circle
centred on 1 (50ohms) with a radius of 4. i.e. on a constant VSWR circle.

Now if we change the source impedance to 100ohms and repeat the same sweep
and re-plot, keeping the chart normalized to 50 ohms, the circle moves on
the resistance axis, still with a radius of 4 and now passing though 2 (100
ohms) resistive. The centre moves to about 0.6 (30ohms). It then becomes
obvious that the locus of the circle is NOT a constant VSWR against
frequency.

You will come to the same conclusion if you normalize the chart to 100 ohms,
the new source impedance and re-plot.

The coax is acting as an impedance transformer, causing a shift along the
resistance axis.

Looking at it another way, the vswr changes sinusoidally with frequency, in
our example, between 2:1 and 8:1. (The same as the Smith chart plot with a
circle of radius 4 centred at about 0.6).

73
Jeff