Is there an embedding of any vector space $V$ into $V^*$?
As far as I know it is not true. The statement that I know of is that there is natural embedding of $V$ into $V^{**}$
Is there any condition/ special vector spaces for which $V$ can be into $V^*$ naturally?
On
You have such a thing in spaces with an inner product. A inner product $\langle x,y\rangle$ is an inner product on a vector space $V$ over field $K$, i.e. a map $$ \langle \cdot,\cdot \rangle \,:\, V\times V \to K $$ that is linear in it's first argument, and either linear or conjugate linear (i.e. $\langle x,\lambda y \rangle = \bar\lambda \langle x,y\rangle $) in it's second argument.
If $\langle \cdot, \cdot \rangle$ is bilinear (respectively conjugate linear in it's second argument), then a linear (respectively conjugate linear) embedding of $V \to V^*$ is given by $$ \cdot^* \,:\, V \to V^* \,:\, c \mapsto \left(x \mapsto \langle x,c \rangle \right) $$ If $\langle \cdot, \cdot \rangle$ is bilinear, it's clear that $x^*$ is a linear map $V \to K$ for every $x$, and that $\cdot^*$ is linear as a map $V \to V^*$. For it to be injective, it must hold that $x^*$ is the zero map only if $x$ is the zero vector. This is e.g. the case if $\langle \cdot,\cdot \rangle$ is positive definite, because then for every non-zero $x$, $x^*(x) = \langle x,y \rangle > 0$. For generally, it suffices for the injectivity of $\cdot^*$ that for every $x$ there is an $y$ with $\langle x,y \rangle \neq 0$. If $\langle \cdot, \cdot \rangle$ is no bilinear, but rather conjugate linear in it's second argument, then the construction above yields a conjugate linear map $V \to V^*$.
Conversely, if you have an embedding $\cdot^* \,:V \to V^*$, you can define an inner product by $$ \langle \cdot,\cdot \rangle \,:\, V\times V \to K \,:\, \langle x,y \rangle = y^*(x) \text{.} $$ Note, however, that in the general case this inner product might not be positive definite, and not (conjugate) symmetric.
If we drop all pretense of naturality, talk pure vector spaces only, i.e. $V^*$ means the algebraic dual of $V$, and assume the axiom of choice (AC), we can do the following:
Let $V$ be a vector space over $K$, and $B$ be a basis of $V$ - the existence of which is guaranteed by AC. Then, by the very definition of vector space basis, there's a coordinatization map $$ \mathfrak{C}_B \,:\, V \to K^B \,:\, x \mapsto (x_b)_{b \in B} $$ such that for all $x \in V$ only finitely many $x_b \neq 0$ and $$ x = \sum_{b \in B} x_b b \text{.} $$ Note that $\mathfrak{C}_B$ is linear, if we interpret sums and products on $K^B$ pointwise, i.e. we have that $\left(\mathfrak{C}_B(x+y)\right)_b = \left(\mathfrak{C}_B(x)\right)_b + \left(\mathfrak{C}_B(y)\right)_b$ and also $\left(\mathfrak{C}_B(\lambda x)\right)_b = \lambda\left(\mathfrak{C}_B(x)\right)_b$ for all $b \in B$.
We can now define a inner product on $V$ just as we do in the finite-dimensional case, i.e. set $$ \langle \cdot,\cdot \rangle \,:\, V\times V \to K \,:\, \langle x,y \rangle = \sum_{b \in B} x_b \overline{y_b} \quad\text{where $(x_b)_{b \in B} = \mathfrak{C}_B(x)$ and $(y_b)_{b \in B} = \mathfrak{C}_B(y)$.} $$ (If the underlying field is not $\mathbb{C}$, replace $\overline{y_b}$ with just $y_b$). That sum is well-defined because if only finitely many $x_b\neq 0$ and $y_b \neq 0$, then also only finitely many summands $x_by_b \neq 0$. The linearity of $\langle \cdot,\cdot \rangle$ in both argument follows from the linearity of $\mathfrak{C}_B$. If the underlying field $K$ is $\mathbb{R}$ or $\mathbb{C}$ this inner product is positive definite, because then for $x \neq 0$ you have $\sum_{b\in B} x_b\overline{x_b} = \sum_{b \in B} |x_b|^2 > 0$. If, on the other hand, $K = \textrm{GF}(p^n)$, then $\langle x,x \rangle = 0$ if $x = b_1 + \ldots b_n$ for $n$ distinct $b_1,\ldots,b_n \in B$.
And now, just as above, we have an embedding $V \to V^*$, defined by $$ \cdot^* \,:\, V \to V^* \,:\, c \mapsto \left(x \mapsto \langle x,c \rangle \right) $$ If $x \in V$, $x \neq 0$ and $(x_b)_{b \in B} = \mathfrak{C}_B(x)$ is the coordinatization of $x$, then $x_{b} \neq 0$ for some $b$. Then $x^*(b) = x_b \neq 0$, i.e for every non-zero vector $x$, $x^*$ is not the zero map, which proves that $\cdot^*$ is injective (even if the inner product was not positive definite).
If $V$ has an inner product $\langle \cdot,\cdot\rangle$, then for each $v\in V$ we can construct an element $f_v\in V^*$ by $$ f_v(x) = \langle x,v\rangle \quad x\in X. $$ This yields a natural embedding of $V$ in $V^*$ for inner product spaces.