The convex conjugate also known as Legendre–Fenchel transformation of a convex function $f:\mathbb{R}^n\to\mathbb{R}\cup\{+\infty\}$ is definite by $$ f^{\ast}(x^\ast)=\sup_{x\in\mathbb{R}^n}\{\langle x,x^\ast\rangle -f(x)\} $$
Question. For what functions, $\varphi:\mathbb{R}\times\mathbb{R}\cup\{+\infty\}\to \mathbb{R}\cup\{+\infty\}$ and $\Phi: \mathbb{R}^n\to \mathbb{R}^n$ known a priori there are functions $\psi: \mathbb{R}\times\mathbb{R}\cup\{+\infty\}\to \mathbb{R}\cup\{+\infty\}$ and $\Psi:\mathbb{R}^n\to \mathbb{R}^n$ such that $$ \Big\lgroup \varphi\, \Big(\, x \, , \,f\circ \Phi\,(\,x\,)\, \Big) \Big\rgroup^{\ast} = \psi \Big( x^\ast,\,f^{\ast}\circ\Psi\,(\,x^\ast)\, \Big) ? $$
In my efforts, I found in wikipedia a table below with some transformations that answer my question for some very particular cases. To be more precise in what I am looking for a theorem which schematically says more or less the following.
Theorem. Let a convex function $f:\mathbb{R}^n\to \mathbb{R}\cup \{\infty\}$ ( not identically equal to infinity). Let functions $\varphi: \mathbb{R}\times \mathbb{R}\cup \{\infty\}\to \mathbb{R}\cup \{\infty\} $ and $\Phi: \mathbb{R}^n\to \mathbb{R}\cup \{\infty\}$. Supose that
Assumption 1: $\varphi$ satisfies the following suitable conditions...
Assumption 2: $\Phi$ satisfies the following suitable conditions...
Then there are functions $\psi: \mathbb{R}\times\mathbb{R}\cup\{+\infty\}\to \mathbb{R}\cup\{+\infty\}$ and $\Psi:\mathbb{R}^n\to \mathbb{R}^n$ such that $$ \Big\lgroup \varphi\, \Big(\, x \, , \,f\circ \Phi\,(\,x\,)\, \Big) \Big\rgroup^{\ast} = \psi \Big( x^\ast,\,f^{\ast}\circ\Psi\,(\,x^\ast)\, \Big)$$
In case $n = 1$ the first four lines of the table give a positive answer to my question under the condition that $\Phi$ and $\varphi$ are afim functions, that is,
$$
\varphi (x,y) = \alpha +\beta \cdot x+ \gamma \cdot y
\qquad
\Phi (x)= \lambda \cdot x + \delta \qquad \gamma >0
$$
then there are functions $\psi$ and $\Psi$ as above such that
$$
\psi(x^\ast,y^\ast )= -\alpha -\delta\cdot\frac{x^\ast-\beta}{\lambda}\
+\lambda \cdot y^\ast
\qquad
\Psi(x^\ast)=\frac{x^\ast -\beta}{\lambda\cdot \gamma}
$$
