Note that there always is a stupid answer. For any $T$ the Dirac delta at $0$ will do the trick. However there are operators $T$ for which there does not exist a measure with nonzero variance.

Try the operator

$$ Tg(x)=g'(x)-xg(x). $$

Suppose that there is a probability measure $\mu$ associated to it such that $\newcommand{\bE}{\mathbb{E}}$ $\DeclareMathOperator{\var}{var}$ $$ \bE_\mu\big[ Xg(X)\big]= \var_\mu(X)\bE_\mu\big[ Tg(X)\big]. $$

$\newcommand{\si}{\sigma}$
Set $\si^2:=\var_\mu[X]$. We deduce $$(1+\si^2)\bE_\mu[Xg(X)\big]=\si^2\bE_\mu[g'(X)],\;\;\forall g. $$

If we take $g(x)=1$ we deduce $\bE[X]=0$.If we take $g(x)=x$ we deduce

$$ (1+\si^2)\si^2=(1+\si^2)\bE_\mu[X^2]=\si^2\bE_\mu[1]=\si^2 $$

which shows that $\si^2$ so $\mu$ must be the Dirac delta concentrated at the origin.

Note that there always is a stupid answer. For any $T$ the Dirac delta at $0$ will do the trick. However there are operators $T$ for which there does not exist a measure with nonzero variance.

Try the operator

$$ Tg(x)=g'(x)-xg(x). $$

Suppose that there is a probability measure $\mu$ associated to it such that $\newcommand{\bE}{\mathbb{E}}$ $\DeclareMathOperator{\var}{var}$ $$ \bE_\mu\big[ Xg(X)\big]= \var_\mu(X)\bE_\mu\big[ Tg(X)\big]. $$

$\newcommand{\si}{\sigma}$
Set $\si^2:=\var_\mu[X]$. We deduce $$(1+\si^2)\bE_\mu[Xg(X)\big]=\si^2\bE_\mu[g'(X)],\;\;\forall g. $$

If we take $g(x)=1$ we deduce $\bE[X]=0$.If we take $g(x)=x$ we deduce

$$ (1+\si^2)\si^2=(1+\si^2)\bE_\mu[X^2]=\si^2\bE_\mu[1]=\si^2 $$

which shows that $\si^2=0$ so $\mu$ must be the Dirac delta concentrated at the origin.

Note that there always is a stupid answer. For ant $T$ the Dirac delta at $0$ will do the trick. However there are operators $T$ for which there does not exist a measure with nonzero variance.

Try the operator

$$ Tg(x)=g'(x)-xg(x). $$

Suppose that there is a probability measure $\mu$ associated to it such that $\newcommand{\bE}{\mathbb{E}}$ $\DeclareMathOperator{\var}{var}$ $$ \bE_\mu\big[ Xg(X)\big]= \var_\mu(X)\bE_\mu\big[ Tg(X)\big]. $$

$\newcommand{\si}{\sigma}$
Set $\si^2:=\var_\mu[X]$. We deduce $$(1+\si^2)\bE_\mu[Xg(X)\big]=\si^2\bE_\mu[g'(X)],\;\;\forall g. $$

If we take $g(x)=1$ we deduce $\bE[X]=0$.If we take $g(x)=x$ we deduce

$$ (1+\si^2)\si^2=(1+\si^2)\bE_\mu[X^2]=\si^2\bE_\mu[1]=\si^2 $$

which shows that $\si^2$ so $\mu$ must be the Dirac delta concentrated at the origin.

Note that there always is a stupid answer. For any $T$ the Dirac delta at $0$ will do the trick. However there are operators $T$ for which there does not exist a measure with nonzero variance.

Try the operator

$$ Tg(x)=g'(x)-xg(x). $$

Suppose that there is a probability measure $\mu$ associated to it such that $\newcommand{\bE}{\mathbb{E}}$ $\DeclareMathOperator{\var}{var}$ $$ \bE_\mu\big[ Xg(X)\big]= \var_\mu(X)\bE_\mu\big[ Tg(X)\big]. $$

$\newcommand{\si}{\sigma}$
Set $\si^2:=\var_\mu[X]$. We deduce $$(1+\si^2)\bE_\mu[Xg(X)\big]=\si^2\bE_\mu[g'(X)],\;\;\forall g. $$

If we take $g(x)=1$ we deduce $\bE[X]=0$.If we take $g(x)=x$ we deduce

$$ (1+\si^2)\si^2=(1+\si^2)\bE_\mu[X^2]=\si^2\bE_\mu[1]=\si^2 $$

which shows that $\si^2$ so $\mu$ must be the Dirac delta concentrated at the origin.