= Solution
Define the <free algebra functor> $F^T$ by $F^T X=(TX,\mu_X)$ and $F^T f=Tf$. The <monad> laws make $\mu_X$ an algebra action, and <naturality> of $\mu$ makes $Tf$ a <monad algebra morphism>. For an <algebra for a monad> $(A,a)$, define
$$
\mathcal C^T(F^TX,(A,a))\longrightarrow\mathcal C(X,A),\quad h\longmapsto h\eta_X,
$$
with proposed inverse $g\mapsto aTg$. This inverse is a <monad algebra morphism>, since
$$
(aTg)\mu_X=a\mu_A T^2g=aTa\,T^2g=aT(aTg).
$$
One composite is $aTg\eta_X=a\eta_Ag=g$. For a <monad algebra morphism> $h$, the other is $aT(h\eta_X)=aTh\,T\eta_X=h\mu_X T\eta_X=h$. Both formulas commute with <composition in a category> in $X$ and with <monad algebra morphisms> in $(A,a)$, so the <bijection> is natural. Thus
$$
\boxed{F^T\dashv U.}
$$
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