This is in some sense a continuation of this problem.
Given a group $G$ I would like to exhibit two actions of $G$ on a set $[n] =\{1,\ldots,n\}$ such that the two actions are isomorphic yet not equivalent.
To recall we say that two group actions $\alpha,\beta : G \mapsto S_n$ are isomorphic if there is an automorphism $f:G \mapsto G$ and bijection $\varphi:[n] \mapsto [n]$ such that
$$\varphi(o^{\alpha(g)}) = \varphi(o)^{\beta(f(g))} \quad \mbox{ for all } o \in [n], g \in G.$$
Finally, we say that they are equivalent if $f$ is the identity map.
The question is as the title says
Given a group $G$ can we find two isomorphic but not equivalent actions of $G$?
I guess not much can be done if $G$ is the trivial group, but what if not?
Your allowance of the bijection $\varphi$ is the same as allowing for an inner automorphism of $S_n$. In other words, you can rephrase your definition of isomorphic actions as:
Two actions (ie, homomorphisms) $\alpha,\beta : G\rightarrow S_n$ are isomorphic if there is an automorphism $f\in\text{Aut}(G)$ and an element $\sigma\in S_n$ corresponding to the inner automorphism $c_\sigma\in\text{Aut}(S_n)$ such that $\beta = c_\sigma\circ\alpha\circ f$. (Note that every automorphism of $S_n$ for $n\ne 2,6$ is inner, so we'll assume that $n\ne 2,6$ from now on)
Your question is then to ask: if $\alpha,\beta$ are isomorphic, then is $\beta = \alpha\circ f'$ for some $f'\in\text{Aut}(G)$.
Note that if the images of $\alpha,\beta$ are distinct, then of course it's impossible for them to be equivalent. On the other hand, they could be isomorphic. For example, if the image of $G$ is not normal in $S_n$, and $\sigma\in S_n$ an element which does not normalize $\alpha(G)$, then $\beta = c_\sigma\circ\alpha$ is isomorphic to $\alpha$ but not equivalent.
So lets assume $\alpha(G) = \beta(G)$.
If $\beta = c_\sigma\circ\alpha\circ f$ and $\sigma\in\alpha(G)$, then you can pick a preimage $g\in \alpha^{-1}(\sigma)$, in which case $\beta = \alpha\circ c_g\circ f$, where $c_g\in\text{Aut}(G)$ is just conjugation by $g$, so you can pick $f' = c_g\circ f$.
In general the answer is no. For example, you could embed $(\mathbb{Z}/5\mathbb{Z})^2$ inside $S_{25}$ via its action on itself. Let $G = \mathbb{Z}^2$, and consider the maps $\alpha,\beta : \mathbb{Z}^2\rightarrow(\mathbb{Z}/5\mathbb{Z})^2$ given by $\alpha$ sending $(x,y)\mapsto(x,y)\mod 5$, and $\beta$ sending $(x,y)\mapsto(2x,y)$.
Note that $\text{Aut}(\mathbb{Z}^2)\cong GL_2(\mathbb{Z})$ - in particular every automorphism has determinant $\pm 1$. On the other hand, the unique automorphism $\gamma$ of $(\mathbb{Z}/5\mathbb{Z})^2$ sending $(x,y)\mod 5\mapsto (2x,y)\mod 5$ has determinant $\equiv 2\mod 5$, which does not come from any automorphism of $\mathbb{Z}^2$ (since those have determinants $\pm 1$ and $2\not\equiv\pm 1\mod 5$).
Finally note that because $(\mathbb{Z}/5\mathbb{Z})^2$ is embedded in $S_{25}$ via the regular representation, the automorphism $\gamma$ extends to an (inner) automorphism of $S_{25}$ (the proof is very easy, but if you just want validation of this fact, see, for example, https://mathoverflow.net/questions/119377/automorphism-of-finite-groups-and-hurwitz-spaces). Thus, $\alpha,\beta$ are isomorphic ($\beta = \gamma\circ\alpha$) but not equivalent (there is no $f\in\text{Aut}(\mathbb{Z}^2)$ such that $\beta = \alpha\circ f$).