= Solution
Identify the <circle group> with $[0,1)$ modulo its endpoints, and let $\lambda$ be normalized <Lebesgue measure>. We first check invariance. On the $r$th inverse branch, $x=(y+r)/m$ for $0\leq r<m$. For an integrable <function> $g$, substitution gives
$$
\int_0^1g(E_mx)\,dx
=\sum_{r=0}^{m-1}\frac1m\int_0^1g(y)\,dy
=\int_0^1g(y)\,dy.
$$
In particular, testing <indicator functions> shows that the <integer multiplication map on the circle> preserves $\lambda$.
To prove <ergodicity>, let $B$ be a measurable set with $\lambda(E_m^{-1}B\mathbin{\triangle}B)=0$ and put $h=\mathbf1_B$. Then $h(E_mx)=h(x)$ almost everywhere. Its <Fourier coefficients> are
$$
c_k=\int_0^1h(x)e^{-2\pi ikx}\,dx\qquad(k\in\mathbb Z).
$$
Using invariance of $h$ and substituting separately on the same $m$ branches gives
$$
\begin{aligned}
c_k
&=\int_0^1h(E_mx)e^{-2\pi ikx}\,dx\\
&=\frac1m\sum_{r=0}^{m-1}e^{-2\pi ikr/m}
\int_0^1h(y)e^{-2\pi i(k/m)y}\,dy\\
&=\begin{cases}0,&m\nmid k,\\c_{k/m},&m\mid k.\end{cases}
\end{aligned}
$$
In the last line, the finite <geometric series> of $m$th <roots of unity> is zero unless $m$ divides $k$, when it equals $m$. For any nonzero integer $k$, divide repeatedly by $m$ until the resulting integer is no longer divisible by $m$. The recurrence then implies $c_k=0$.
Since the <Fourier basis> is a complete <orthonormal basis> of $L^2([0,1),\lambda)$, the absence of every nonconstant <Fourier coefficient> implies $h=c_0=\lambda(B)$ almost everywhere. An <indicator function> takes only the values zero and one, so $\lambda(B)$ must be zero or one. This proves
$$
\boxed{E_m\text{ is ergodic for Lebesgue measure for every integer }m\geq2.}
$$
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