I tried converting Ito's lemma to Stratonovich form, but I got inconsistent results.
Consider a 1-D SDE in both Ito and Stratonovich sense:
$$
dX_t = \mu(X_t) dt + \sigma(X_t) dW_t
= \bar{\mu}(X_t) dt + \sigma(X_t) \circ dW_t,
$$
where
$$
\bar{\mu} = \mu – \frac{1}{2}\sigma \sigma'. \qquad(1)
$$
Let $f=f(x)$ be a smooth function of $x$. For simplicity, $f, \mu$, and $\sigma$ have no time dependence. By Ito's lemma, we get
$$
df =
\left( \mu f' + \frac{1}{2} \sigma^2 f'' \right) dt + \sigma f' dW_t.
$$
Now, I want to transform the above equation into Stratonovich form, but I got inconsistent results. Here are my derivations.
1. Using transformation formula
From Eq.(1), we have:
$$
\begin{align}
df &=
\left[
\mu f' + \frac{1}{2} \sigma^2 f''
– \frac{1}{2} \sigma f' \frac{\partial (\sigma f')}{\partial X}
\right] dt + \sigma f' \circ dW_t\\
&=
\left[
\mu f' + \frac{1}{2} \sigma^2 f''
– \frac{1}{2} \sigma \sigma' (f')^2 – \frac{1}{2} \sigma^2 f' f''
\right] dt + \sigma f' \circ dW_t.
\qquad \qquad (2)
\end{align}
$$
2. Using the chain fule for Stratonovich SDE
Since Stratonovich SDE follows the regular chain rule, we have
$$
\begin{align}
df
&= f' \circ dX_t
= f' (\bar{\mu} dt + \sigma\circ dW_t ) \\
&= \left(
\mu f' – \frac{1}{2} \sigma \sigma' f'
\right) dt + \sigma f' \circ dW_t.
\qquad \qquad (3)
\end{align}
$$
It seems like these two results can't be identical, but I'm not sure where it goes wrong.
Edit: Corrected a few typos according to Kurt's comment.
Best Answer
In this answer I show that for any process $Y_t$ whose stochastic integrals are defined we have $$ \tag{1} \underbrace{\int_0^tY_s\circ\,dW_s}_{\text{Stratonovich}}=\underbrace{\int_0^tY_s\,dW_s}_{\text{Ito}}+\tfrac{1}{2}\langle W,Y\rangle_t\, $$ (the notation for $Y_s$ was $f(s)$ in the link).
In your case we have an Ito SDE $$ dX_t=\mu\,dt+\sigma\,dW_t $$ and we know \begin{align}\tag{Ito} df(X_t)&=f'(X_t)\,dX_t+\tfrac 12f''(X_t)\,d\langle X\rangle_t\\[2mm] &=f'(X_t)\sigma(X_t)\,dW_t+f'(X_t)\mu(X_t)\,dt+\tfrac 12f''(X_t)\sigma^2(X_t)\,dt\,.\tag{2} \end{align} Claim: $$\tag{3} df=f'\left(\mu-\frac{\sigma\sigma'}{2}\,dt\right)+f'\sigma\circ dW_t\,. $$ In particular, if $\mu\equiv 0$ and $\sigma\equiv 1$ then $X_t$ is a Brownian motion and the Stratonovich calculus follows the rules of ordinary calculus: $$\tag{Stratonovich} \boxed{\quad df(W_t)=f'(W_t)\circ\,dW_t\,.\phantom{\Big|}\quad} $$ Proof. Writing $$ \alpha(X_t)=\sigma(X_t)f'(X_t) $$ we get from Ito $$ d\alpha=\alpha'\,dX_t+\frac{\alpha''}{2}\,d\langle X\rangle_t\,. $$ Using $Y_t=\alpha(X_t)$ and (2) this means $$\tag{4} d\langle Y,W\rangle_t=\alpha'(X_t)\,\sigma(X_t)\,dt=\Big(\sigma\,\sigma'\,f'+\sigma^2\,f''\Big)(X_t)\,dt\,. $$ Using now (1), (2) and (4) we get (3). $$\tag*{$\Box$} \quad $$ To answer the question where your mistake was:
In short: in your equation (2) you have too many $f'\,.$
Up to a typo you have correctly from Ito $$\tag{5} df=\left(\mu f'+\frac12\sigma^{\color{red}2}f''\right)\,dt+\sigma f'\,dW_t\,. $$
When going to $\sigma f'\circ dW_t$ you follow the pattern $$ \mu\,dt+\sigma\,dW_t=\left(\mu-\frac{\sigma\sigma'}{2}\right)\,dt+ \sigma\circ dW_t $$ which is correct for $dX_t$ but not for $df(X_t)\,.$
This incorrect approach leads to \begin{align}\tag{incorrect} df&=\left(\mu f'+\frac12\sigma^2 f''-\frac{\sigma f'(\sigma f')'}2\right)\,dt+\sigma f'\circ dW_t\\ &=\left(\mu f'+\frac12\sigma^2 f''-\frac{\sigma\sigma'(f')^\color{red}{2}+\sigma^2f' f''}2\right)\,dt+\sigma f'\circ dW_t\\ \end{align} where $\sigma^2$-terms don't cancel and the $\sigma\sigma'$ term has one $f'$ too many.
The correct step from (5) to (3) is to use the expression (4) which accounts via (1) for another drift change when going from $\sigma f'dW_t$ to $\sigma f'\circ dW_t\,.$
I have to say that this was new and surprising to me. Every time I look at Stratonovich I learn something new.