Conjugation Quandles and... "Quandle-Groups"? From quandles to Groups This question is already asked MathSE
A quandle $(Q,*,/ )$ is a idempotent right-distributive and right invertible structure. 

1) $a*a=a$
2) $(a*b)*c=(a*c)*(b*c)$
3) $(a*b) /b=(a/b)*b=a$

If we have a group $(G, \cdot, e,^{-1})$ and $*$ is the cojugation operation on $G$
$$a*b:=bab^{-1}$$
and 
$$a/b:=a*b^{-1}$$
then $(G,*,/)$ is denoted with $Conj(G)$ and is a quandle because it satisfies the quandles axioms

1) $a*a=a=aaa^{-1}$
2) $(a*b)*c=(a*c)*(b*c)$ 

because $c(bab^{-1})c^{-1}=(cbc^{-1})cac^{-1}(cb^{-1}c^{-1})=cbab^{-1}c^{-1}$

3) $(a*b) *b^{-1}=(a*b^{-1})*b=a$

because $b^{-1}(bab^{-1})b=b(b^{-1}ab)b^{-1}=a$
I read that we have too that a group homomorphism between two groups $G$ and $G'$ is a quandle homomorphism between theire cojugation quandles $Conj(G)$ and $Conj(G')$ and that makes $Conj$ a functor betwen the category of Groups and the category of quandles...

I wanted to know more about this functor $Conj$ that "maps" Groups to Quandles and since I'm not expert of category theory I apologize if I use a wrong terminology

$Q1a$ - I learnt that not every Quandle is a conjugation Quandle or in
  other words $conj$ is not ""surjective"" on the "set" of all quandles
  (that is not a set but a class i think) so how can I prove that a
  Quandle is a Conjugation Quandle too? 
$Q1b$ -Wich extra "axioms" must hold in a Quandle that is a
  Conjugation Quandle?

This has something to do with the inverse construction of $Conj$ so my next question is

$Q2$ - There is a way to define a group operation starting with a
  quandle operation? Like an inverse $Conj$ construction that build a "Quandle-Group" $Conj^{-1}(Q)$ from a conjugation quandle $Q$.
$Q3$ Is this process not unique?

With not unique I mean that is possible to have two different groups $G=(G,\cdot,\phi)$ and $G'=(G,\circ,\varphi)$ and $$Conj (G)= Conj(G')$$
this should mean that is possible to have 
$$b\cdot a\cdot \phi(b)=b \circ a \circ \varphi(b)$$
where $\phi(b)$and $\varphi(b)$ are the inverse functions and
$$a \cdot b \neq a \circ b$$

$Q4$ My last question is if possible to generalize the conjugation
  operation of groups for monoids and semigruops in a way that these
  "Monoid-conjugations" and "Semigroup-conjugations" are quandles. If is possible I would like to read more about.

 A: You have two excellent expert answers. Here is a very simple answer to some of your questions.
First a slight overview. Given a group one has the conjugation operation $a*b=bab^{-1}$ This is just $a*b=a$ when $G$ is an abelian group. A group with $n$ elements gives a quandle with $n$ elements which satisfies the axioms you listed. 
A finite quandle can be thought of as an $n$ by $n$ "multiplication" table which satisfies certain rules including every symbol appears once in each column and : in the $x$ column, the symbol $x$ occurs in the $x$ row. I'm not sure how interesting quandles are without a group or perhaps another structure to motivate them.
Anyway, there is only one group of order $5$ and it is abelian so it has quandle 
$$\left[ \begin {array}{ccccc} 0&0&0&0&0\\ 1&1&1&1&1
\\ 2&2&2&2&2\\3&3&3&3&3
\\ 4&4&4&4&4\end {array} \right] 
.$$
This is the only quandle with $5$ elements which can come from a group.
Here are two other quandles with $5$ elements $$\left[ \begin {array}{ccccc} 0&4&3&2&1\\ 2&1&0&4&3
\\ 4&3&2&1&0\\ 1&0&4&3&2
\\ 3&2&1&0&4\end {array} \right] $$
$$ \left[ \begin {array}{ccccc} 0&3&1&4&2\\ 3&1&4&2&0
\\ 1&4&2&0&3\\ 4&2&0&3&1
\\ 2&0&3&1&4\end {array} \right] $$
We can see that they are not isomorphic since only the second has $a*b=b*a$. It might not be hard to find all quandles (up to isomorphism) with $5$ elements, but I did not do that.
The axioms could be checked by brute force but it is easier if one knows that the first is $a*b=2a-b \mod 5$ and the second $a*b=3a-2b \mod 5$. The general construction for $m$ elements is $a*b=ka-(k-1)b \mod m$ where we need that $k$ is relatively prime to  $m.$ 
In answer to a later question: The construction is actually very general. Suppose we have a ring (maybe not commutative) with an identity, and that $k$ is an element with a multiplicative inverse $k^{-1}$. Then the operation $a*b=ka+(1-k)b$ certainly satisfies $a*a=a.$ Also $$(a*c)*(b*c)=$$ $$k(ka+(1-k)c)+(1-k)(kb+(1-k)c)=k^2a+(1-k)kb+(1-k)c$$ while $$(a*b)*c=k(ka+(1-k)b)+(1-k)c$$ so these are equal, since in all cases $(1-k)k=k(1-k).$ So it remains to see what is needed to have a solution $c=a/b$ to $c*b=a.$ That would have to be $a/b=k^{-1}(a+(k-1)b)$  in which case $(a*b)/b=k^{-1}(ka+(1-k)b+(k-1)b)=a.$ 
A: Let me point out a sort of characterization of conjugation quandles. (This is too long for a comment.) 
Let $X$ be a finite quandle. The enveloping group $G_X$ is the group with generators $x\in X$ and relations $x(x*y)=yx$ for all $x,y\in X$. 
The enveloping group has the following universal property: 

For any group $G$ and any map $f:X\to G$ satisfying
  $f(x*y)=f(x)^{-1}f(y)f(x)$ there exists a unique group homomorphism
  $g:G_X\to G$ such that $f=g\circ\partial$, where $\partial:X\to G_X$,
  $i\mapsto x_i$.

See for example:


*

*Andruskiewitsch, Nicolás; Graña, Matías. From racks to pointed Hopf algebras. Adv. Math. 178 (2003), no. 2, 177--243. MR1994219 (2004i:16046)

*Fenn, Roger; Rourke, Colin. Racks and links in codimension two. J. Knot Theory Ramifications 1 (1992), no. 4, 343--406. MR1194995 (94e:57006), link
Now suppose that $X$ is connected. (This means that the group generated by the
permutations $\varphi_x\colon y\mapsto y*x$, $x\in X$, acts transitively on
$X$.) Then it follows from the definitions that all the permutations $\varphi_x$ have the same order. Suppose that the order of the permutations $\varphi_x$ is $d$. Then:


*

*$\delta(x)^d=\delta(y)^d$ for all $x,y\in X$,

*$\delta(x)^d$ is central in $G_X$, and

*the quotient group $F_X=G_X/\langle \delta(x)^d\rangle$ is finite.


Since there is a group homomorphism $\deg:G_X\to\mathbb{Z}$ given by $\deg(x)=1$
for all $x\in X$, the group $G_X$ is $\mathbb{Z}$-graded. Now
consider the quotient group $F_X$ and let  $p:G_X\to F_X$ be the canonical
surjection. Write the generators of $G_X$ as $x_i$ for all $i\in X$. 
The following is useful to check whether a connected quandle is a conjugacy class:

The map $\partial:X\to G_X$ is injective if and only if the map $X\to F_X$,
  $i\mapsto p(x_i)$, is injective.

This follows from the following observation: $p$ restricted to the conjugacy class of $x_1$ is bijective. To prove this, let $u,v$ be conjugate elements of $G_X$. Then $\deg(u)=\deg(v)$. If $p(u)=p(v)$ then $u=vx_1^m$ for some $m$. Taking degrees, $m=0$ and hence $u=v$.
Application.
This website contains computational results on small connected quandles and their knot colorings obtained by W. Edwin Clark and Timothy Yeatman. In this subpage you can find a list of 16 quandles which are not known to be conjugation quandles. Apparently it is not easy to use the computer and the universal property of $G_X$ to check whether these quandles are conjugation quandles. The group $F_X$ does the trick and one can easily decide which of these 16 quandles are conjugation quandles!
A: The notion of a semigroup with conjugation can be formalized by the notions of an LD-monoid, LRD-monoid, and LRDQ-monoid. The information in this answer can be found in the book Braids and Self-Distributivity by Patrick Dehornoy.
$\textbf{LD-monoids}$
An LD-monoid (LD stands for left-distributive) is an algebra $(M,\cdot,1,\wedge )$ such that $(M,\cdot,1)$ is a monoid and the following identities are satisfied:
$$x\cdot y=(x\wedge y)\cdot x,(x\cdot y)\wedge z=x\wedge (y\wedge z),
x\wedge (y\cdot z)=(x\wedge y)\cdot(x\wedge z),1\wedge x=x,x\wedge 1=1.$$
If $(M,\cdot,1,\wedge )$ is an LD-monoid, then $(M,\wedge )$ is generally not the quandle. However, $(M,\wedge )$ still satisfies the left-distributivity law
$x\wedge (y\wedge z)=(x\wedge y)\wedge (x\wedge z)$. Algebras that satisfy the law
$x\wedge (y\wedge z)=(x\wedge y)\wedge (x\wedge z)$ are commonly called LD-systems and LD-systems are still of interest to at least some knot theorists even though they are usually not quandles..
The operation $\wedge$ acts as a sort of conjugation operation on monoids.
For instance, suppose that $(M,\cdot,1)$ is a monoid where for all $x,y\in M$ there exists a unique $z$ where $z\cdot x=x\cdot y$ (in particular, if $M$ is a group). Then $(M,\cdot,1)$ becomes an LD-monoid when we attach the operation $x\wedge y$ defined by $(x\wedge y)\cdot x=x\cdot y.$ Furthermore, if $(G,\cdot,1,\wedge)$ is an LD-monoid and $G$ is a group, then $x\wedge y=x\cdot y\cdot x^{-1}$.
One can even construct LD-monoids by taking a collection of functions and a sort of conjugacy operation. For example, let $\mathfrak{I}_{\infty}$ be the collection of all injective functions from $\mathbb{N}$ to $\mathbb{N}$. Then $(\mathfrak{I}_{\infty},\cdot,1,\wedge)$ is an LD-monoid where $\cdot$ is the composition function, $1$ is the identity mapping, and $(f\wedge g)(x)=f\circ g\circ f^{-1}(x)$ whenever $x\in Im(f)$ and where $(f\wedge g)(x)=x$ otherwise. 
$\textbf{LRDQ-monoids and quandles}$
Even though LD-monoids do not give us quandles, one can generalize the notion of an LD-monoid to LRDQ-monoids and LRDQ-monoids do give us quandles.
An LRD-monoid is an algebra $(M,\cdot,1,\wedge,\vee)$ that satisfies the following identities. 
$$1\wedge x=1\vee x=x,x\wedge 1=x\vee 1=1;x\cdot y=(x\wedge y)\cdot x=y\cdot(y\vee x)$$
$$(x\cdot y)\wedge z=x\wedge(y\wedge z),(x\cdot y)\vee z=y\vee(x\vee z)$$
$$x\wedge(y\cdot z)=(x\wedge y)\cdot(x\wedge z),x\vee(y\cdot z)=(x\vee y)\cdot(x\vee z).$$
An LRDQ-monoid is an LRD-monoid that satisfies the identity $x\wedge(x\vee y)=x\vee(x\wedge y)=x$. Every group becomes an LRDQ-monoid with operations $\wedge$ and $\vee$ defined by $x\wedge y=xyx^{-1},x\vee y=x^{-1}yx$.
If $(M,\cdot,1,\wedge,\vee)$ is an LRDQ-monoid, then $\wedge$ is idempotent if and only if $\vee$ is idempotent. If $(M,\cdot,1,\wedge,\vee)$ is an LRDQ-monoid such that $\wedge$ and $\vee$ are idempotent, then $(X,\wedge,\vee)$ is a quandle.
For example, suppose that $(M,\cdot,1)$ is a monoid such that for all $x,y$ there are
unique $z_{L},z_{R}$ such that $xy=z_{L}x=xz_{R}$. Then one may define unique operations $\wedge$ and $\vee$ on $(M,\cdot,1)$ such that $x\cdot y=(x\wedge y)\cdot x$ and $x\cdot y=y\cdot(y\vee x)$. Then $(M,\cdot,1,\vee,\wedge)$ is an LRDQ-monoid, and the operations $\vee$ and $\wedge$ are idempotent, so $(M,\vee,\wedge)$ is a quandle.
LD-monoids originally to axiomatize the identities satisfied by the algebra of elementary embeddings obtained from the I3 axiom which states that there exists some non-trivial elementary embedding $V_{\lambda}\rightarrow V_{\lambda}$. The collection of elementary embeddings from $V_{\lambda}$ to $V_{\lambda}$ becomes an LD-monoid. The I3 axiom is one of the strongest large cardinal axioms that far extends the axioms of set theory.
A: Q1b Joyce in A classifying invariant of knots, the knot quandle. J. Pure Appl. Algebra 23 (1982) proves that the equational theory of quandles and that of groups endowed with a quandle structure via conjugation is the same. So conjugation quandles are not a subvariety of quandles. There is however one more axiom which is necessary (I follow your convention of right quandle action)
$$ y * x = y \iff x *y =x $$
Q1a I'm not aware of any sufficient condition for a quandle to be a conjugation subquandle of a group.
Q2 Given a quandle you can construct its Enveloping Group, or Adjoint Group, which is the free group on the elements modulo the relations 
$$y *x = x^{-1}y x $$
Joyce shows that this construction is left adjoint to that of taking the conjugation quandle. This is more explicitly stated in  Andruskiewitsch, Nicolás; Graña, Matías From racks to pointed Hopf algebras. Adv. Math. 178 (2003)
Q3 Consider that every abelian group with the same number of elements give rise to the same (trivial) quandle
