= Solution
Write the <monad> identities as
$$
\mu_A\eta_{TA}=1_{TA}=\mu_AT\eta_A,\qquad
\mu_AT\mu_A=\mu_A\mu_{TA}.
$$
If $\mu_A$ is invertible, its two right inverses agree, so $\eta_{TA}=T\eta_A$. Conversely, suppose $\eta_T=T\eta$. <Naturality> of $\eta$ at $\mu_A$, followed by that equality at $TA$, gives
$$
\eta_{TA}\mu_A=T\mu_A\eta_{TTA}=T\mu_AT\eta_{TA}=T(\mu_A\eta_{TA})=1_{TTA}.
$$
Together with $\mu_A\eta_{TA}=1_{TA}$ this proves
$$
\boxed{\mu\text{ is invertible}\ \Longleftrightarrow\ \eta_T=T\eta,\qquad\mu^{-1}=\eta_T=T\eta.}
$$
This is the equivalence defining an <idempotent monad>.
For the <equalizer submonad of a monad>, let $\beta_A:RA\to TA$ equalize $\eta_{TA}$ and $T\eta_A$. If $f:A\to B$, <naturality> of these two transformations shows that $Tf\beta_A$ equalizes the corresponding pair at $B$. There is therefore a unique $Rf$ satisfying
$$
\beta_BRf=Tf\beta_A.
$$
Uniqueness gives $R1=1$ and $R(gf)=Rg\,Rf$, so $R$ is a <functor> and $\beta:R\Rightarrow T$ is a <natural transformation>.
Take the stipulated factorizations $\alpha_A$ and $\theta_A$. <Naturality> of $\beta$ at $\beta_A$ rewrites their defining equations as
$$
\beta_A\alpha_A=\eta_A,\qquad
\beta_A\theta_A=\mu_A\beta_{TA}R\beta_A=\mu_AT\beta_A\beta_{RA}.
$$
The composites on the right are natural in $A$; cancellation of the <monic arrow> $\beta_A$ therefore makes $\alpha$ and $\theta$ natural. We now verify every <monad> law, rather than assuming that restricting the structure automatically preserves them.
For the two unit laws, <naturality> and the unit laws of $T$ give
$$
\begin{aligned}
\beta_A\theta_A\alpha_{RA}
&=\mu_AT\beta_A\eta_{RA}
=\mu_A\eta_{TA}\beta_A=\beta_A,\\
\beta_A\theta_AR\alpha_A
&=\mu_AT\beta_AT\alpha_A\beta_A
=\mu_AT\eta_A\beta_A=\beta_A.
\end{aligned}
$$
Cancelling $\beta_A$ proves $\theta_A\alpha_{RA}=1_{RA}=\theta_AR\alpha_A$. For associativity, the same <naturality> equations give
$$
\begin{aligned}
\beta_A\theta_AR\theta_A
&=\mu_AT\mu_A\,TT\beta_A\,T\beta_{RA}\,\beta_{RRA},\\
\beta_A\theta_A\theta_{RA}
&=\mu_A\mu_{TA}\,TT\beta_A\,T\beta_{RA}\,\beta_{RRA}.
\end{aligned}
$$
These are equal by associativity of $\mu$; cancelling $\beta_A$ proves associativity of $\theta$. Thus
$$
\boxed{(R,\alpha,\theta)\text{ is a monad, and }\beta:R\Rightarrow T\text{ is a monad morphism}.}
$$
The remaining claims use two <split equalizers associated with a monad>. First, $\eta_{TA}:TA\to TTA$ is a <split equalizer> of $\eta_{TTA},T\eta_{TA}$. Its retraction is $\mu_A$, and the splitting of the parallel pair is $T\mu_A$, since
$$
\mu_A\eta_{TA}=1,\qquad
T\mu_A\eta_{TTA}=\eta_{TA}\mu_A,\qquad
T\mu_AT\eta_{TA}=1.
$$
The equalizing equation follows from <naturality> of $\eta$. In general, identities $re=1$, $sf=er$, $sg=1$ and $fe=ge$ prove that $e$ is an <equalizer>: if $fh=gh$, then $h=erh$, with unique factorization $rh$. Hence $\eta_{TA}$ and $\beta_{TA}$ are two <equalizers> of the same pair. Their comparison is $\alpha_{TA}$, and its inverse is $\mu_A\beta_{TA}$. In particular
$$
\boxed{\alpha_{TA}\text{ is invertible},\qquad \alpha_{TA}^{-1}=\mu_A\beta_{TA}.}
$$
Second, $e_A=T\eta_A:TA\to TTA$ is a <split equalizer> of $TT\eta_A,T\eta_{TA}$, with retraction $\mu_A$ and splitting $\mu_{TA}$. Indeed,
$$
\mu_AT\eta_A=1,\qquad
\mu_{TA}TT\eta_A=T\eta_A\mu_A,\qquad
\mu_{TA}T\eta_{TA}=1.
$$
This is precisely the parallel pair in the PDF's diagram after $T\beta_A$. The map $T\beta_A$ equalizes that pair, because applying $T$ to the defining <equalizer> equation gives their equality on $T\beta_A$. Put
$$
h_A=\mu_AT\beta_A:TRA\to TA.
$$
The split-equalizer factorization and $\beta_A\alpha_A=\eta_A$ give
$$
e_Ah_A=T\beta_A,\qquad h_AT\alpha_A=1_{TA}.
$$
If $T\alpha_A$ is invertible, then $T\beta_A=e_A(T\alpha_A)^{-1}$ is a <monic arrow>. Conversely, if $T\beta_A$ is a <monic arrow>, the equation
$$
T\beta_A(T\alpha_Ah_A)=e_Ah_A=T\beta_A
$$
forces $T\alpha_Ah_A=1_{TRA}$, so $h_A$ is the inverse of $T\alpha_A$. Therefore
$$
\boxed{T\alpha_A\text{ invertible}\ \Longleftrightarrow\ T\beta_A\text{ monic}.}
$$
Notice that, under either condition, $T\beta_A$ is actually a <split monomorphism>, with retraction $T\alpha_A\mu_A$. Thus $TT\beta_A$ is a <monic arrow> too, since a <functor> preserves <split monomorphisms>. We have not assumed that $T$ preserves arbitrary <monomorphisms>.
Now prove the specified <naturality> square is a <pullback in a category>. Let $x:X\to TRA$ and $y:X\to RTA$ satisfy $T\beta_Ax=\beta_{TA}y$. <Naturality> of $\eta$, followed by the <equalizer> equation for $\beta_{TA}$, gives
$$
\begin{aligned}
TT\beta_A\eta_{TRA}x
&=\eta_{TTA}T\beta_Ax
=\eta_{TTA}\beta_{TA}y\\
&=T\eta_{TA}\beta_{TA}y
=T\eta_{TA}T\beta_Ax
=TT\beta_AT\eta_{RA}x.
\end{aligned}
$$
Cancel the <monic arrow> $TT\beta_A$. Thus $x$ equalizes $\eta_{TRA},T\eta_{RA}$ and factors uniquely as $x=\beta_{RA}z$ with $z:X\to RRA$. <Naturality> of $\beta$ then gives
$$
\beta_{TA}R\beta_Az=T\beta_A\beta_{RA}z=T\beta_Ax=\beta_{TA}y.
$$
Cancel $\beta_{TA}$ to obtain $R\beta_Az=y$. This is exactly existence and uniqueness in the pullback <universal property>.
Because the square is a pullback and $T\beta_A$ is a <monic arrow>, $R\beta_A$ is a <monic arrow>. <Naturality> of $\alpha$ and the formula for $\theta$ now give
$$
R\beta_A\alpha_{RA}\theta_A
=\alpha_{TA}\beta_A\theta_A
=\alpha_{TA}\mu_A\beta_{TA}R\beta_A
=R\beta_A.
$$
Cancelling $R\beta_A$ proves $\alpha_{RA}\theta_A=1_{RRA}$. The other composite is already one of the unit laws. Hence $\theta_A$ is invertible. If the equivalent monicity conditions hold for every $A$, we conclude
$$
\boxed{(R,\alpha,\theta)\text{ is idempotent},\qquad \theta^{-1}=\alpha_R=R\alpha.}
$$
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