What do you mean by `algebraic units of modulus 1 have modulus ≠1'? I believe that we may have roots with modulus 1 but not roots of unity.

I meant $M$ is irreducible. Anyway, I have just taken more time to check, and the formula I initially guesses is wrong, since the density of $T(\mu)$ with regard to $T(\pi)$ is more complicated that what I initially though.

I think that the answer will be negative. I assume that $T$ is irreducible, and I call $\pi$ the invariant probability measure. Then my impression is that $$\frac{dT(\mu)}{d\mu(x)} = \frac{\mu(x_2)/\pi(x_2)}{\mu(x_1)/\pi(x_1)}.$$ If this formula is correct, the density takes only finitely many values.

The sum in the formula giving $\mu[x_1=n_1;\ldots;x_k=n_k]$ should be replaced by a product. If you assume that $[\nu_1,\ldots,\nu_n] M = [\nu_1,\ldots,\nu_n]$ (namely the initial distribution is stationary), the Markov shift operator preserves $\mu$. I think that you should assume only that the measure $\nu M$ is equivalent to $\nu$.

Are the two Brownian bridges independent? If yes, there difference is $\sqrt{2}$ times a Brownian bridge, so the distribution is the same as $2\int_0^1 B_t^2 dt$, where $B$ is a Brownian bridge.

@Nate River. I completed the proof.

@Iosif Pinellis I added a sentence to explain why the two measures are mutually singular.

@Christian Remling The main difference is that I do not require $g$ to be infinite at 0.

@Giorgo Metafune I tried to build $g$ with the help of $h(x)=x\ln(x)$, but I did not succeed yet. I found that the error term in the difference between the integral and the Riemann sum was too small, but I am not sure of my computations. Adapting Euler Mac Laurin techniques to get estimates is not very simple.