This is not a full answer, but perhaps too long for a comment.
Your question is about the distance problem in a fixed Cayley graph. If one also considers the Cayley graph as part of the input then some things are known. This makes sense in the context of permutation groups, where we can explicitly give the generators in a succinct way.
Even and Goldreich in the paper "Minimum-Length Generator Sequence Problem is NP-Hard" (J. ALGORITHMS. Vol. 2, no. 3, pp. 311-313. 1981) showed:
We examine the following questions:
(1) Given a set of generators of a
permutation group G and a target
permutation P, find (the length of) a
shortest generator sequence realizing
P. (2) Given a set of generators of a
permutation group G, find the minimum
upper bound on the length of generator
sequences needed to realize any
permutation in G. We show that both
problems are NP-Hard by reducing the
3XC problem to each of them. The
reductions we use show that these
results hold even if the given set of
generators is restricted to contain
for each generator its inverse, too.
This actually makes the problem NP-Complete (when the length is written in unary).
In other work, Mark Jerrum in "The complexity of finding minimum-length generator sequences" (Theoretical Computer Science Volume 36, 1985, Pages 265-289) showed:
The computational complexity of the
following problem is investigated:
Given a permutation group specified as
a set of generators, and a single
target permutation which is a member
of the group, what is the shortest
expression for the target permutation
in terms of the generators? The
general problem is demonstrated to be
Image -complete and, indeed,is shown
to remain so even when the generator
set is restricted to contain only two
permutations. The restriction on
generator set cardinality is the best
possible, as the problem becomes
soluble in polynomial time if the
generator set contains only one
permutation. An interesting feature of
this problem is that it does not fall
under the headings of ‘two person
games’ or ‘formal languages’ which
cover the great majority of known
Image -complete problems. Some
restricted versions of the problem, in
which the generator set is fixed
rather than being part of the problem
instance, are also investigated and
shown to be computationally tractable.
One result of this kind is that
determining the most compact
expression of a permutation in terms
of ‘cyclicly adjacent transpositions’
can be achieved in polynomial time.
Thus, from an initial arrangement of
distinct objects on a circle, one can
quickly compute the smallest number of
interchanges of adjacent objects
required to realise any other
arrangement. Surprisingly, this
problem appears substantially more
difficult to solve than the related
one (for which a solution has been
known for some time) in which the
objects are arranged on a line
segment.
Note that Jerrum considers the length of the path in binary, which changes things.
Best Answer
The shortest (in terms of weight) path, constrained to have exactly n (or at most n) edges, can be found in polynomial time. For instance, given your graph $G=(V,E)$ make an expanded graph $H$ that has as its vertices the pairs $(v,i)$ where $v\in G$ and $0\le i\le n-1$. Draw an edge in $H$ from $(v,i)$ to $(w,i+1)$ whenever $G$ has an edge from $v$ to $w$, with the same weight. Then the shortest $n$-edge path in $G$ from $v$ to $w$ is the same as the shortest path in $H$ from $(v,0)$ to $(w,n)$. To look for paths in $G$ that are at most $n$ edges long, add to $H$ edges of weight zero from $(v,i)$ to $(v,i+1)$.
However, these paths allow repeated vertices and edges. If repetitions are disallowed, and $G$ has $n+1$ vertices, then the shortest length-$n$ path is just a Traveling salesman path, so of course it's NP-complete.