By definition, an automorphism is an isomorphism from $G$ to $G$, while an isomorphism can have different target and domain.
In general (in any category), an automorphism is defined as an isomorphism $f:G \to G$.
In simple terms, two graphs are isomorphic to each other so long as there is a bijection between the two vertex sets and such that the bijection preserves edges. That is, two graphs $G$ and $G'$ are isomorphic if there exists a bijection, $\phi: V(G) \to V(G')$, and if that bijection also preserves edges. Explicitly, this means that if $uv$ is an edge in $E(G)$, then $\phi (u) \phi (v)$ is an edge in $E(G')$.
One can visualize a graph isomorphism in two ways.
1. Start with a graph and move around vertices in what ever way you want while keeping all the edges in tact. The result, a graph that looks different, but isn't really.
2. A re-labeling of the vertices of $G$. This produces a graph that looks the same, but the vertices are called something else.
Automorphisms are a little bit trickier. The way I look at an automorphism is a moving around of vertices (while keeping edges in tact, as in (1) above) with the caveat that the graph must look exactly the same as before. For example, take $C_4$ with vertex set $V(G) = u_1,u_2,u_3,u_4$ and edge set $E(G) = u_1u_2,u_2u_3,u_3u_4,u_4u_1$ and consider the following bijection.
1. $\phi: V(G) \to V(G')$ where $\phi$ is the permuation $(u_1u_2u_3u_4)$ that sends $u_1 \to u_2$ $u_2 \to u_3$ $u_3 \to u_4$, and $u_4 \to u_1$. (If one is unfamiliar with cycle notation). This is an isomorphism since every edge is preserved, and indeed it is also an automorphism since the resulting graph looks exactly the same as the regular graph. (the permuation above is really just a rotation by 90 degrees if you lay out the original as a "square", as is tradition.
Best Answer
While you can label the vertices of a path graph in $n!$ different ways, most of of these labelings do not correspond to isomorphisms between the graphs. For example, if the vertex labelled by $1$ in the first graph were one an end of the path but was not on the end in the second graph, any map that sent vertex $1$ in the first graph to vertex $1$ in the second graph would not be an isomorphism since isomorphisms cannot map a vertex in the first graph to a vertex in the second graph that has a different degree. So from your $n!$ ways of labeling the graphs, you obtain $n!$ graphs which are all isomorphic, but there are just two isomorphisms between any given pair of these graphs.
In fact, the number of isomorphisms between two isomorphic graphs is always equal to the number of automorphisms of one of the graphs. This is because if $X$ and $Y$ are graphs, $\mathrm{Aut}(X)$ is the automorphism group of $X$ and $\phi : X \rightarrow Y$ is an isomorphism, then the coset of all isomorphisms from $X$ to $Y$ is $\phi \circ \mathrm{Aut}(X) = \{\phi \circ \sigma \vert \sigma \in \mathrm{Aut}(X)\}$. Clearly, this set has the same cardinality as $\mathrm{Aut}(X)$. (Actually, this last paragraph holds not just for graphs but for just about any mathematical object you can imagine.)
To answer the question in your edit, this does not depend on whether the two graphs have the same vertex set.