Is there any theorem on uniqueness of integrating factor for inexact ordinary differential equations? Do we have any theorem regarding  uniqueness of integrating factor for inexact ordinary differential equations?
Anything like:
If $f(v)$ and $g(v)$ are functions of $v$ and integrating factors for the given inexact ODE, then $f(v)=g(v)$
 A: I found this example in Boyce and DiPrima. The differential equation $$3xy+y^2+(x^2+xy){dy\over dx}=0$$ has the integrating factor $\mu(x)=x$, but also the integrating factor $\mu(x,y)=\bigl(xy(2x+y)\bigr)^{-1}$. 
A similar example is given here: the differential equation $$(4y^2+3xy)dx-(3xy+2x^2)dy=0$$ has the integrating factors $${1\over xy(x+y)}\qquad{\rm and}\qquad{y\over x^5}$$
Another example is given in Michael D Greenberg's text, Ordinary Differential Equations: $2y\,dx+3x\,dy=0$ has integrating factors $x^{-1/3}$ and $(xy)^{-1}$. Of course, this example is separable, and the integrating factor $(xy)^{-1}$ is just separating the variables. Indeed, the differential equation $ay\,dx+bx\,dy=0$ will have integrating factors $x^{(a-b)/b}$, $y^{(b-a)/a}$, and $(xy)^{-1}$. 
EDIT: Here is an example in line with OP's additional request in the comments. 
Let $P(x,y)=-x^2y^{-3}$, $Q(x,y)=xy^{-2}$. Then $(y/x)^n$ is an integrating factor for all integer $n$. 
MORE EDIT: This may be more what you want: If $\mu^m$ and $\mu^n$ are both integrating factors for the same differential equation, and $m\ne n$, then the equation was already exact to begin with. 
Proof. We suppose $(1)\quad(\mu^mP)_x=(\mu^mQ)_y$,$\qquad$ $(2)\quad(\mu^nP)_x=(\mu^nQ)_y$,$\quad$ and $m\ne n$. From (1) we get $\quad(3)\quad\mu^mP_x+m\mu^{m-1}\mu_xP=\mu^mQ_y+m\mu^{m-1}\mu_yQ$, which is $(4)\ \ \mu P_x+m\mu_xP=\mu Q_y+m\mu_yQ$. Similarly, from (2) we get $\ (5)\ \ \mu P_x+n\mu_xP=\mu Q_y+n\mu_yQ$. Multiply (4) by $n$, multiply (5) by $m$, and subtract, to get $(n-m)\mu P_x=(n-m)\mu Q_y$. Since $m\ne n$, this implies $P_x=Q_y$, so the original equation was exact. 
EVEN MORE EDIT: If $\mu$, $\nu$, and $\mu\nu$ are all (nonzero) integrating factors for a differential equation, then the equation was exact to begin with. 
Proof. We assume $\mu$, $\nu$, and $\mu\nu$ are all integrating factors, so $(\mu P)_x=(\mu Q)_y$, $(\nu P)_x=(\nu Q)_y$, and $(\mu\nu P)_x=(\mu\nu Q)_y$. Expanding, we get $$\eqalign{\mu P_x+\mu_xP&=\mu Q_y+\mu_yQ\cr\nu P_x+\nu_xP&=\nu Q_y+\nu_yQ\cr\mu\nu P_x+\mu\nu_xP+\mu_x\nu P&=\mu\nu Q_y+\mu\nu_yQ+\mu_y\nu Q}$$ Multiply the second equation by $\mu$ and subtract from the third equation to get $$\mu_x\nu P=\mu_y\nu Q$$ Divide that equation by $\nu$ and subtract from the first equation to get $\mu P_x=\mu Q_y$, so $P_x=Q_y$, and the original equation was exact. 
A: If $$0=dF=F_xdx+F_ydy$$ defines a direction field, then for any differentiable and monotonous $\phi$, $\phi'\ne 0$, by the chain rule
$$0=d(\phi\circ F)=\phi'(F)F_xdx+\phi'(F)F_ydy$$
defines the same direction field.

$\mu$ is an integrating factor for $0=Mdx+Ndy$ so that $dF=\mu(Mdx+Ndy)$, that is, $F_x=\mu M$, $F_y=\mu N$, then $g(F)\mu$ is also an integrating factor (on some connected subset where $g(F)\ne 0$), with $\phi=G$ an anti-derivative of $g$.

In the first example of the other answer, the integrating factor $\mu=x$ gives the first integral $$F=x^3y+\frac12x^2y^2=\frac12x^2y(2x+y)$$ and the second integrating factor is $\frac{\mu}{2F}$
