(This solution is based on the idea of geodude).
Let $M$ be a manifold with a connection $\nabla$, and let $A \in C^\infty (M)$.
Define another connection: $\tilde \nabla_X Y = \nabla_X Y + X(A)Y.$
We need two technical lemmas (proofs are provided below).
Lemma $\text{1}$: $R^{\nabla} = R^{\tilde \nabla}$
Lemma $\text{2}$: Let $\phi \in \text{Iso}(M,\nabla)$. Then $\phi \in \text{Iso}(M,\tilde \nabla) \iff dA_q \circ d\phi_{\phi^{-1}(q)} = dA_{\phi^{-1}(q)}$.
Here is the idea:
Since $\phi \in \text{Iso}(M,\nabla) \Rightarrow \phi \in \text{Iso}(M,R^{\nabla}) \stackrel{\mathrm{Lemma 1}}{=} \text{Iso}(M,R^{\tilde \nabla})$, it is enough to find an isomorphism of $\nabla$, which is not an isomorphism of $\tilde \nabla$. By Lemma 2, we have an explicit condition which is easy to violate, thus finding a suitable isormophism as required.
A concrete example:
$M= \mathbb{R},\nabla $ the Levi-Civita connection w.r.t the standard metric on $\mathbb{R}$, $\phi(q) = -q = \phi^{-1}(q)$ is an isometry, hence it preserves $\nabla$. For a given function $A: \mathbb{R} \rightarrow \mathbb{R}$, $\phi$ preserves $\tilde \nabla^A$ iff $-dA_q = dA_{-q} \iff A'(-q)=-A'(q)$, but there are many functions $A$ which do not satisfy this.
Proof of Lemma $\text{2}$:
$\tilde \nabla_{\phi_*X}({\phi_*Y})=\phi_* (\tilde \nabla_X Y) \iff \nabla_{\phi_*X}({\phi_*Y}) + [(\phi_*X)(A)] \cdot\phi_*Y = \phi_* (\nabla_X Y) + \phi_* (X(A)Y) \stackrel{\mathrm{\phi preserves \nabla}}{\iff} [(\phi_*X)(A)] \cdot\phi_*Y = \phi_* (X(A)Y ) \iff [(\phi_*X)(A)] \cdot\phi_*Y = (X(A)\circ \phi ^{-1}) \cdot\phi_*Y$
Since $Y$ and hence $\phi_*Y$ can be arbitrary vector fields on $M$, and in particular for every $q \in M, \left(\phi_*Y \right)(q)$ can be a non-zero vector in $T_qM$ , this forces $(\phi_*X)(A) = X(A)\circ \phi ^{-1}$.
So $\phi \in \text{Iso}(M,\tilde \nabla) \iff \forall q \in M,X \in \Gamma(TM) , (\phi_*X)(A)(q) = X(A)\circ \phi ^{-1}(q) \iff dA_q((\phi_*X)(q)) = dA_{\phi^{-1}(q)}\left( X(\phi^{-1}(q)) \right ) \iff dA_q\left(d\phi_{\phi^{-1}(q)}(X\left(\phi^{-1}(q)\right)) \right) = dA_{\phi^{-1}(q)}\left( X(\phi^{-1}(q)) \right )$.
Since $X$ is arbitrary, this forces: $dA_q \circ d\phi_{\phi^{-1}(q)} = dA_{\phi^{-1}(q)}$.
Proof of Lemma $\text{1}$:
$\tilde R(X,Y)\,Z = [\tilde\nabla_X,\tilde\nabla_Y]\,Z - \tilde\nabla_{[X,Y]}Z = \tilde\nabla_X (\tilde\nabla_Y Z)-\tilde\nabla_Y (\tilde\nabla_X Z)-\tilde\nabla_{[X,Y]}Z = \tilde\nabla_X (\nabla_Y Z+Y(A)Z)-\tilde\nabla_Y (\nabla_X Z+X(A)Z)-\big[\nabla_{[X,Y]}Z+\big([X,Y]A\big)Z\big] = [\nabla_X (\nabla_Y Z+Y(A)Z) + X(A)(\nabla_Y Z+Y(A)Z)]-[\nabla_Y (\nabla_X Z+X(A)Z)+Y(A)(\nabla_X Z+X(A)Z)]-\big[\nabla_{[X,Y]}Z+\big([X,Y]A\big)Z\big] = $
$ R(X,Y)Z + \nabla_X(Y(A)Z) + X(A)(\nabla_Y Z+Y(A)Z) - \nabla_Y (X(A)Z)-Y(A)(\nabla_X Z+X(A)Z)-\big([X,Y]A\big)Z$
So,
$\tilde R(X,Y)\,Z - R(X,Y)Z = X(Y(A))Z+Y(A)\nabla_X Z + X(A)(\nabla_Y Z+Y(A)Z) - Y(X(A))Z - X(A) \nabla_Y (Z)-Y(A)(\nabla_X Z+X(A)Z)-\big([X,Y]A\big)Z = 0.$