For question one: Just knowing that two groups act simply transitively on the same set set isn't enough to give you a canonical isomorphism between them; you still must pick an element of the set to get the isomorphism. This is equivalent to choosing a path between the two basepoints, which is the old-school way to get a (noncanonical) iso on $\pi_1$s with varying basepoints.
Here is how it goes.
Let $B$, be a space nice enough to have a (simply connected) universal cover, say $B$ is connected, locally connected and semi-locally simply connected. Let $(X,x_0)\to (B,b_0)$ be its universal cover.
Take a loop $\gamma: (S^1,1)\to (B,b_0)$ then you can lift $\gamma$ to a path $\overline{\gamma}: I\to X$ that projects to $\gamma$. Now $\overline{\gamma}(1)$ is an element of $X_{b_0}$. You can use then the following theorem.
Let $(Y,y_0)\to (B,b_0)$ be a (path) onnected and locally path connected space over $B$ and $(X,x_0)\to (B,b_0)$ is a cover of $B$, then a lift of $(Y,y_0)\to (B,b_0)$ to $(Y,y_0)\to (X,x_0)$ exists iff the image of $\pi_1(Y,y_0)$ inside $\pi_1(B,b_0)$ is contained in the image of $\pi_1(X,x_0)$ inside $\pi_1(B,b_0)$
Use the previous theorem with $(Y,y_0)=(X,\overline{\gamma}(1) )$.
This tells you that there exists a covering map $X\to X$ sending $x_0$ to $\gamma(1)$.
It is easy to see that this map depends only on the homotopy class of $\gamma$ using the following result
Let $(X,x_0)$ be a cover of $(B,b_0)$ and $Y$ be a connected space over $B$. If two liftings of $Y\to B$ to $Y\to X$ coincide at some $y_0$ in $Y$, the they're equal.
This tells you that if $\overline{\gamma}(1)=\overline{\tau}(1)$ then the two morphisms $X\to X$ you get, coincide.
Moreover, using the inverse of $\gamma$, you see that the morphisms $X\to X$ you get are automorphisms.
This gives you a well defined map $\pi_1(B,b_0)\to \text{Aut}_B(X)$.
Using what I said before, it is easy to see that it is an isomorphism.
Best Answer
Almost always, these are two different actions.
The monodromy action is by path lifting, lifting a given path at each of the points in the fiber and mapping each such point to the endpoint of the corresponding lift. This makes sense for any covering space $q:(Y,y_0) \to (X,x_0)$, where $Y$ is possibly disconnected.
To get a well-defined action by the deck group $G=G(\widetilde{X})$, which can be identified with $\pi_1(X,x_0)$ in the universal cover $p:(\widetilde{X},\widetilde{x}_0) \to (X,X_0)$, on the fiber of an intermediate connected normal cover $q:(Y,y_0) \to (X,x_0)$, certain elements need to identified: $G(Y) \cong G/N$, where $N= q_*\pi_1(Y,y_0)$. So, there is an action of $G$ on $F = q^{-1}(x_0)$ that factors through the action of $G/N$ on $F$.
Here's an example where these two actions are different:
Let $X = S^1 \vee S^1$ with oriented labeled edges $x$ and $y$, identified with generators of $\pi_1(X)$. And let $Y = \widetilde{X}$ be its universal cover, viewed as a graph (4-valent tree) with labeled oriented edges and its vertex set identified with $\pi_1(X)$.
Consider the pair of adjacent vertices $(1,y)$. The monodromy action by the element $x$ sends this pair to $(x,yx)$, which are not adjacent in $Y$. The deck group action by the element $x$ sends the pair to $(x,xy)$.
In general, the deck group action extends to a homeomorphism of $Y$. But the monodromy action need not extend continuously.
The above example is related to other actions: if $Y$ is a normal covering graph with fundamental group $N \trianglelefteq G$, where $G$ is a free group, then $Y$ is the right Cayley graph of the presentation $G/N$. The action of $F/N$ by deck transformations is the left action (by graph automorphisms). The monodromy action is the action on the vertex set by right multiplication (which does not extend to an action by graph automorphisms).
Exercise #27 in Chapter 1 of Hatcher's Algebraic Topology book is worth a look.