Consider a theory of fields $\phi:M\to T$ where $M$ is a manifold, and $T$ is a set. In physics, $T$ is often either a vector space or a manifold. We call $M$ the domain of the theory, and we call $T$ the target space. of the theory. We call a function from $M$ to $T$ a field configuration, and the set of all field configurations is denoted $\mathcal F$.
Let's consider two situations:
Case 1. Let groups $G_M$ and $G_T$ be given. Let $\rho_M$ be an action of $G_M$ on $M$, and let $\rho_T$ be an action of $G_T$ on $T$, then there is a "natural" action of $\rho_\mathcal F$ of $G_M\times G_T$ on $\mathcal F$ given by
\begin{align}
\rho_\mathcal F(g_M, g_T)(\phi)(x) = \rho_T(g_T)\Big(\phi\big(\rho_M(g_M)^{-1}(x)\big)\Big)
\end{align}
I'll leave it to you to prove that this is a group action.
Case 2. Let a groups $G$ be given. Let $\rho_M$ be an action of $G$ on $M$, and let $\rho_T$ be an action of $G$ on $T$, then there is a "natural" action of $\rho_\mathcal F$ of $G$ on $\mathcal F$ given by
\begin{align}
\rho_\mathcal F(g)(\phi)(x) = \rho_T(g)\Big(\phi\big(\rho_M(g)^{-1}(x)\big)\Big)
\end{align}
I'll leave it to you to prove that this is in fact a group action.
Now, let us phrase your question in the following way:
In either of the above cases, is their a sense in which $\rho_M$ induces $\rho_T$?
The answer, as far as I am aware, is that it depends on the context and on what you mean by "induced." Let's consider the example you give in your statement of the question.
Example. An $\mathrm{O}(2)$ vector field on $\mathbb R^{3,1}$.
We have
\begin{align}
M = \mathbb R^{3,1}, \qquad T = \mathbb R^2, \qquad G_M = \mathrm{P}(3,1), \qquad G_T = \mathrm{O}(2)
\end{align}
where $\mathrm{P}(3,1)$ is the Poincare group in four dimensions. In this case, one often takes $\rho_M$ and $\rho_T$ to be
\begin{align}
\rho_M(\Lambda,a)(x) = \Lambda x+a, \qquad \rho_T(R)(v) = Rv
\end{align}
Notice that this falls under Case 1 above. In this case, there is no "canonical" relationship (as far as I am aware) between the Poincare group and $\mathrm{O}(2)$, so there is no canonical sense in which $\rho_M$ induces $\rho_T$.
However, consider the following example:
Example. An $\mathrm{SO}(3)$ $2$-tensor field on $\mathbb R^{3}$.
We have
\begin{align}
M = \mathbb R^{3}, \qquad T = T_2(\mathbb R^3), \qquad G = \mathrm{SO}(3)
\end{align}
where $T_2(\mathbb R^3)$ is the vector space of $2$-tensors on $\mathbb R^3$. Then there is a natural action of $G$ on $M$ given by
\begin{align}
\rho_M(R)x = Rx
\end{align}
Since $2$-tensors on $\mathbb R^3$ can be described as bilinear maps functions $S:\mathbb R^3\times \mathbb R^3 \to \mathbb R$, there is also a natural action of $G$ on $T$ given by
\begin{align}
\rho_T(S)(x_1, x_2) = S(R^{-1}x_1, R^{-1} x_2)
\end{align}
which can also be written as
\begin{align}
\rho_T(S)(x_1, x_2) = S(\rho_M(R)^{-1}x_1, \rho_M(R)^{-1} x_2)
\end{align}
so, in this case, there is a sense in which $\rho_M$ has induced $\rho_T$.
