Inducing a Lie algebra action from a Lie group action

Let $$G$$ be a Lie group, $$M$$ be a (smooth) manifold and assume that $$G$$ acts smoothly on $$M$$ with a fixed point $$p \in M$$. (I mean, there is a Lie group homomorphism $$\rho : G \to C^{\infty}(M,M)$$ where $$C^{\infty}(M,M)$$ is a mapping space endowed with (possibly infinite dimensional) smooth manifold structure and each element $$\rho(g) : M \to M$$ fixes $$p$$.)

Since $$\rho(g)(p) = p$$ for all $$g \in G$$, we can consider the action $$G \to \mathrm{GL}(T_{p}G)$$ on the tangent space at $$p$$. Then the tangent map of this action at the identity gives a Lie algebra homomorphism $$\mathfrak{g} \to \mathrm{End}(T_{p}G)$$.

However, we can take the tangent map of $$\rho$$ directly and obtain $$\mathfrak{g} \to T_{\mathrm{Id}}C^{\infty}(M,M)$$. My question is, is there any canonical map from $$T_{\mathrm{Id}}C^{\infty}(M,M)$$ to $$\mathrm{End}(T_{p}G)$$, so that the two constructions above coincides in some sense?

The intuition behind your question does not really work out. The problem is that the isotropy representation of $$G$$ is already obtained by differentiating the map $$x\mapsto \rho(g)(x)$$ in the point $$p$$. Only after that, you have to differentiate the resulting map on $$G$$ in order to obtain the isotropy representation of $$\mathfrak g$$. In your proposed alternative way to obtain this, you only consider the differentiation of the map on $$G$$ but not of the one on $$M$$.
Second, there is the technical problem that $$C^\infty(M,M)$$ is certainly not a Lie group, you would have to take the diffeomorphism group $$Diff(M)$$ here. Form compact $$M$$ this is indeed an infinite dimensional Lie group with Lie algebra the space $$\mathfrak X(M)$$ of vector fields on $$M$$. Already here things are not easy technically since you need manifolds modelled on Frechet spaces. In the non-compact case, things become significantly more complicated.
However, you don't need infinite dimensional manifolds to get the infinitesimal version of a group action: The action of $$G$$ on $$M$$ gives rise to a Lie algebra (anti-)homomorphism $$\mathfrak g\to M$$ described by the so-called fundamental vector fields. Given $$X\in\mathfrak g$$, you define $$\zeta_X\in\mathfrak X(M)$$ by $$\zeta_X(x):=\tfrac{d}{dt}|_{t=0}\rho(\exp(tX))(x)$$. It is easy to see (using only finite dimensional arguments) that $$\zeta_X$$ is indeed a smooth vector field on $$M$$ and how this map is compatible with Lie brackets. Lie's second fundamental theorem shows that this is indeed a good infinitesimal version of a smooth group action.
However, as you can see from the definition, if $$p\in M$$ is a fixed point of $$G$$, then $$\zeta_X(p)=0$$ for any $$X\in\mathfrak g$$ (which is precisely the infinitesimal version of $$p$$ being a fixed point). To get the isotropy representation of $$\mathfrak g$$ from this, I believe you have to differentiate (in $$M$$ directions). Choosing a linear connection $$\nabla$$ on $$TM$$, you can consider the map $$T_pM\to T_pM$$ defined by $$\xi\mapsto \nabla_{\xi}\zeta_X(p)\in T_pM$$. Since $$\zeta_X(p)=0$$, this mapping is independent of the choice of $$\nabla$$ and it should give the action of $$X$$ on $$\xi$$ under the isotropy representation of $$\mathfrak g$$.