Let $1$ be the unit of $C$. For every $X$, we define $X^* = Hom(X,1)$. I will assume $C$ is strict closed symmetric monoidal. Further assuming the condition the OP specified, we can show $C$ is rigid.

Let's unpack the additional condition the OP wants to assume. For every $X$, the functors $F(-) = X\otimes -$ and $G(-) = Hom(X,-)$ form an adjoint equivalence. In particular, the counit of the adjunction, $\epsilon: FG \to 1_C$ is a natural isomorphism. So, $\epsilon_1: X\otimes X^* = X\otimes Hom(X,1) \to 1$ is an isomorphism. Call this morphism $ev_X$. Define the coevaluation as its inverse.

Following Section 2.10 of Tensor Categories, we must show that the compositions:

$X \stackrel{coev_X\otimes id}{\longrightarrow} (X\otimes X^*) \otimes X \stackrel{\alpha}{\longrightarrow} X\otimes (X^*\otimes X) \stackrel{id\otimes ev_X}{\longrightarrow} X$, and

$X^*\stackrel{id \otimes coev_X}{\longrightarrow} X^*\otimes (X \otimes X^*) \stackrel{\alpha^{-1}}{\longrightarrow} (X^*\otimes X) \otimes X^* \stackrel{ev_X\otimes id}{\longrightarrow} X^*$

are the identity morphisms. But the first is just $X\cong 1\otimes X$, followed by the associator, followed by $X\otimes 1 \cong X$, which is certainly the identity on $X$ (by strictness), and the second works the same way. So, indeed, $X^*$ is a left dual to $X$.

By symmetry, $X^*$ is also the right dual to $X$, and $X^*\otimes X \cong 1$, so the two morphisms in 2.10.2 of Tensor Categories are also identities. Hence, every $X$ has both a left and right dual, so $C$ is rigid.