Number of Automorphisms of a Irregular Graph. I have been looking for results on  number of graph automorphisms  of irregular graph(upper and lower bound).
 I searched , but could not find anything which can be used directly.
Say, $G$ is $k$ connected irregular graph.
$F_k , f_k $ is the  number of graph automorphisms  of irregular graph $G$, where $n=|G|=$total number of vertices of $G$. 
What is the upper bound $f_k(n)$and lower bound $F_k(n)$?
 A: [EDIT] : the following solution applies to highly irregular graphs (sometimes known as irregular graphs) https://en.wikipedia.org/wiki/Highly_irregular_graph
I'd be surprised if you could find a nice lower bound.  Testing for the irregular graphs on 12 vertices (110 graphs according to Brendan McKay https://cs.anu.edu.au/~bdm/data/graphs.html), most have automorphism group of order 1 or 2 (a single example admits a group of order 4).  All 474 examples on 13 vertices have a trivial automorphism group.
As for an upper bound, you should convince yourself that an automorphism of an (connected) irregular graph that fixes even one vertex must in fact fix the entire graph (why?).  You can consider the number of permutations on $n$ elements that do not fix anything.  But this isn't quite enough, because you need to whole GROUP (all nonidentity elements) to not fix anything! Such a group is said to act semiregularly.
EDIT: After more thinking, since the automorphism group $G$ acts semiregularly, all orbits have size $|G|$.  So $|G|$ divides $n$.  But $G$ can't equal $n$ because that would imply that there is an automorphism sending vertex $a$ to vertex $b$ for all vertex pairs; and we have to send a vertex with degree $d$ to another with degree $d$.  So let $c$ be the number of distinct degrees.  This gives a bound $|G| \leq n/c$.  For example, the special graph on 12 vertices has precisely 4 vertices of degree 1, 4 vertices of degree 2, and 4 vertices of degree 3, arranged perfectly so that there is an automorphism of order 4 = 12/3.
