Rotors are I, II and III. Rings and indicator settings are at their home positions. Reflector is UKW-B. Suppose $E_0$ is an Enigma I machine without a plugboard; at $t=0$, $n_{\text{indicator}}=1$, i.e. $\texttt{A}$. See footnote.
Suppose $E_1$ is the same at $t=1$. We have $E_1 = (\rho^nR\rho^{-n})MLUL^{-1}M^{-1}(\rho^nR^{-1}\rho^{-n})$, where,
$n=n_{\text{indicator}} - n_{\text{ring}} = 1$ (an integer mod 26 in $\mathbb{Z}_{26}$) with $n_{\text{indicator}}=2$, i.e. $\texttt{B}$ and $n_{\text{ring}} = 1$, i.e. $\texttt{01}$;
since $\rho^1 = \rho =
\begin{pmatrix}
\mathtt{ABCDEFGHIJKLMNOPQRSTUVWXYZ} \\
\mathtt{BCDEFGHIJKLMNOPQRSTUVWXYZA}
\end{pmatrix}$,
$\rho^{-1} =
\begin{pmatrix}
\mathtt{ABCDEFGHIJKLMNOPQRSTUVWXYZ} \\
\mathtt{ZABCDEFGHIJKLMNOPQRSTUVWXY}
\end{pmatrix}$ and
$R =
\begin{pmatrix}
\mathtt{ABCDEFGHIJKLMNOPQRSTUVWXYZ} \\
\mathtt{BDFHJLCPRTXVZNYEIWGAKMUSQO}
\end{pmatrix} = \mathtt{(CFLVMZOYQIRWUKXSG)(ABDHPEJT)(N)}$,
we have $\rho R \rho^{-1} = \texttt{(BEKULYNXPHQVTJWRF)(ACGODISZ)(M)}$;
$M =
\mathtt{(FIXVYOMW)(CDKLHUP)(ESZ)(BJ)(GR)(NT)(A)(Q)}$;
$L =
\mathtt{(AELTPHQXRU)(BKNW)(CMOY)(DFG)(IV)(JZ)(S)}$; and
$U =
\mathtt{(AY)(BR)(CU)(DH)(EQ)(FS)(GL)(IP)(JX)(KN)(MO)(TZ)(VW)}$.
Therefore, $E_1 = \texttt{(AB)(CQ)(DM)(EF)(GX)(HI)(JS)(KW)(LP)(NY)(OT)(RV)(UZ)}$.
For example, we can obtain $(\texttt{A})E_1 = \texttt{B}$ from the above expression or via the following diagram.
B A
^ |
| v
rho R rho^-1 = (BEKULYNXPHQVTJWRF)(ACGODISZ)(M)
^ |
| |
+-----------------+|
||
+--------++
| |
v |
M = (FIXVYOMW)(CDKLHUP)(ESZ)(BJ)(GR)(NT)(A)(Q)
| ^
| |
| +----------------+
+------------+ |
| |
v |
L = (AELTPHQXRU)(BKNW)(CMOY)(DFG)(IV)(JZ)(S)
| ^
+----+ |
|+---------------+
v|
U = (AY)(BR)(CU)(DH)(EQ)(FS)(GL)(IP)(JX)(KN)(MO)(TZ)(VW)
Q.E.D.
Footnote:
Indicators $\textbf{n}_{\text{indicator}} = (k_{\text{indicator}}, j_{\text{indicator}}, n_{\text{indicator}})$,
rings $\textbf{n}_{\text{ring}} = (k_{\text{ring}}, j_{\text{ring}}, n_{\text{ring}})$, and
wiring core offsets = $(k, j, n) = \textbf{n}_{\text{indicator}} - \textbf{n}_{\text{ring}}$ where $k, j, n \in \mathbb{Z}_{26} = \{1, 2, \cdots, 25, 0\}$.
$k_{\text{indicator}}, j_{\text{indicator}}, n_{\text{indicator}} \in \mathbb{A} = \{\mathtt{A}, \mathtt{B}, \cdots, \mathtt{Z}\} \cong \mathbb{Z}_{26}$
and
$k_{\text{ring}}, j_{\text{ring}}, n_{\text{ring}} \in \mathbb{B} = \{\mathtt{01}, \mathtt{02}, \cdots, \mathtt{26}\} \cong \mathbb{Z}_{26}$.
P.S. It possibly is not very helpful to keep track of notch positions, which are 8 places ahead of the turnover positions. It is better to use Royal Flags Wave Kings Above or Q-R E-F V-W J-K Z-A.
P.P.S. Expanding the right rotor (III) computation of which you might not be very sure. Turnover position is at V-W. Therefore, there is no middle wheel turnover (mwto) and indicators are $\texttt{AAB}$. $\rho, \rho^{-1}, R$ are elements of a permutation group. For clarity, write $\rho, \rho^{-1}$ in double row notation and $R$ in cycle notation. The latter has three orbits. Their lengths are 17, 8 and 1 respectively.
AB
|^
||
v|
rho = (ABCDEFGHIJKLMNOPQRSTUVWXYZ)
BCDEFGHIJKLMNOPQRSTUVWXYZA
|^
||
++------------------+
| |
++ |
| v
R = (CFLVMZOYQIRWUKXSG)(ABDHPEJT)(N)
^ |
| |
+---+ |
| |
+-+---------------+
| |
| |
v |
rho^-1 = (ABCDEFGHIJKLMNOPQRSTUVWXYZ)
ZABCDEFGHIJKLMNOPQRSTUVWXY
| ^
v |
C E
