Do the forgetful functors $G_H:\bf Monoid \to \bf Semigroup^1$ and $G_O:\bf Semigroup^1 \to \bf Semigroup$ have left and/or right adjoints? Here $\bf Semigroup$ is the category of semigroups with semigroup homomorphisms, $\bf Semigroup^1$ is the category of monoids (semigroups with identity) with semigroup homomorphisms, and $\bf Monoid$ is the category of monoids with monoid homomorphisms (the image of the identity is the identity).
An interesting functor $F_H:\bf Semigroup^1 \to \bf Monoid$ is given by the functor which leaves the monoid objects unchanged, and modifies the homomorphisms by replacing the image of the identity by the identity of the target monoid. Is this a left or right adjoint functor to $G_H$?
A nice functor $F_O:\bf Semigroup \to \bf Semigroup^1$ is the one which leaves monoids unchanged, and adjoins an identity element to semigroups missing an identity element. The homomorphisms are extended by mapping the newly adoint identity element to the identity element of the target monoid. Is this a left or a right adjoint functor to $G_O$, or both?
A well known functor $F:\bf Semigroup \to \bf Monoid$ is the one which adjoins to each semigroup an identity (independent of whether it already has one), and extends the homomorphisms by mapping the newly adoint identity element to the newly adjoint identity element of the target monoid. I have seen the assumption (statement) that this is the left adjoint of the forgetful functor. Is this correct?
The well known functor $F: \bf Semigroup \to \bf Monoid$ is the left adjoint to the forgetful functor $G: \bf Monoid \to \bf Semigroup$. To see that $\hom(F(X),Y) \equiv \hom(X,G(Y))$ natural in the variables $X$ and $Y$, explicitly writing down the natural transformations $\varphi$ and $\psi$ inverse to each other helps.
The compositions $\varphi\circ \psi$ and $\psi \circ \varphi$ are both identities.
If we explicitly spell out the corresponding transformations $\varphi$ and $\psi$ for $F_H$ and $G_H$, then $\psi \circ \varphi$ is an identity, but $\varphi\circ \psi$ is only a projection instead of a true identity (semigroup homomorphism get projected to monoid homomorphisms).
If we explicitly spell out the corresponding transformations $\varphi$ and $\psi$ for $F_O$ and $G_O$, then $\varphi\circ \psi$ is an identity, but $\psi \circ \varphi$ is only a projection instead of a true identity (semigroup homomorphism get projected to monoid homomorphisms) when $X$ is not a monoid.
As Martin Brandenburg wrote in a comment, it is not clear whether $F_H$ and $F_O$ are functors at all. In fact, they fail to be functors, because $F(g\circ f)=F(g)\circ F(f)$ fails.
For $F_H$ look at $\begin{matrix} & f & & g & \\ & & * & \longrightarrow & * \\ * & \longrightarrow & * & \longrightarrow & * \\ * & \longrightarrow & * & {}^{\underline{\quad}\nearrow} \end{matrix}$. We get $\begin{matrix} & F_H(f) & & F_H(g) & \\ & & * & \longrightarrow & * \\ * & {}^{\underline{\quad}\nearrow} & * & \longrightarrow & * \\ * & \longrightarrow & * & {}^{\underline{\quad}\nearrow} \end{matrix}$, so $\begin{matrix} F_H(g) \circ F_H(f)\\ \begin{matrix} * & \longrightarrow & * \\ * & \longrightarrow & * \end{matrix}\end{matrix}$.
But $\begin{matrix} & g \circ f & * \\ * & \longrightarrow & * \\ * & {}^{\underline{\quad}\nearrow} \end{matrix}$, so we get $\begin{matrix} F_H(g \circ f)\\ \begin{matrix} * & \longrightarrow & * \\ * & {}^{\underline{\quad}\nearrow} & * \end{matrix}\end{matrix}$. Hence $F_H(g) \circ F_H(f)$ and $F_H(g \circ f)$ are different.
Here, the vertical stars "describe" specific semilattices (commutative idempotent semigroups), where the result of a multiplication is the lower of the two involved stars (i.e. the meet or the greatest lower bound). Hence the identity element is given by the highest star in each case. The arrows "describe" specific homomorphisms.
For $F_O$ look at $\begin{matrix} & f & & g & * \\ {\scriptsize\begin{matrix}* \quad *\\*\\ \end{matrix}} & \longrightarrow & * & \longrightarrow & * \end{matrix}$. The leftmost semilattice is missing an identity element, hence $F_O$ will adjoint one. We get $\begin{matrix} & F_O(f) & & F_O(g)\\ * & {}_{\overline{\;\cdot\;}\searrow} & & & * \\ {\scriptsize\begin{matrix}* \quad *\\*\\ \end{matrix}} & \longrightarrow & * & \longrightarrow & * \end{matrix}$, so $\begin{matrix} F_O(g) \circ F_O(f) \\ \begin{matrix} * & {}_{\overline{\;\cdot\;}\searrow} & * \\ {\scriptsize\begin{matrix}* \quad *\\*\\ \end{matrix}} & \longrightarrow & * \end{matrix}\end{matrix}$.
But $\begin{matrix} & g \circ f & * \\ {\scriptsize\begin{matrix}* \quad *\\*\\ \end{matrix}} & \longrightarrow & * \end{matrix}$, so we get $\begin{matrix} F_O(g \circ f) \\ \begin{matrix} * & \longrightarrow & * \\ {\scriptsize\begin{matrix}* \quad *\\*\\ \end{matrix}} & \longrightarrow & * \end{matrix}\end{matrix}$. Hence $F_O(g) \circ F_O(f)$ and $F_O(g \circ f)$ are different.