How hard is the Circuit Isomorphism problem?
As pointed out by @Marzio De Biasi, the problem you are asking about is $\mathsf{coNP}$-complete, by a direct reduction from TAUTOLOGY: given a boolean formula $\varphi$, is $\varphi$ equivalent to the constant 1 circuit? It is easily seen to be in $\mathsf{coNP}$. (Given that Marzio really answered the question, I suggest anyone thinking of upvoting this answer flip a coin and if it lands heads then go vote for one of Marzio's other answers.)

Let me also clarify: the problem you are asking about is more typically called Circuit Equivalence. Circuit Isomorphism is the problem: given two circuits $C_1, C_2$ on $n$ inputs, is there a permutation $\pi \in S_n$ such that $C_1(\vec{x})$ and $C_2(\pi(\vec{x}))$ are equivalent? Circuit Isomorphism has a status similar to that of Graph Isomorphism, but one level higher in $\mathsf{PH}$: it is $\mathsf{coNP}$-hard (by the same reduction above), not known to be in $\mathsf{coNP}$, in $\mathsf{\Sigma_2 P}$, but not $\mathsf{\Sigma_2 P}$-complete unless $\mathsf{PH} = \mathsf{\Sigma_3 P}$.