In RSA, one of the math background is:
$m^{φ(n)}\bmod n = 1$, where $m$ is the message to be sent, $n=p*q$.
Yes. More precisely: if $m$ and $n$ are integers with $n>1$ and $\gcd(m,n)=1$, then $m^{φ(n)}\bmod n=1$, where $φ$ is Euler's totient. That's commonly invoked in proof of RSA, including the original RSA article: R.L. Rivest, A. Shamir, and L. Adleman's A Method for Obtaining Digital Signatures and Public-Key Cryptosystem.
The equation is from $λ(n)$ (Carmichael function), which requires $m$ and $n$ are co-primes.
More precisely: $λ(n)$ is the smallest positive integer such that if $m$ and $n$ are integers with $n>1$ and $\gcd(m,n)=1$, then $m^{λ(n)}\bmod n=1$.
since $n=p*q$, thus $φ(n)=λ(n)$, I guess.
No, that's an incorrect conclusion by analogy. If $n$ is an odd composite (as in practice) and not a prime power, then $φ(n)\neλ(n)$, and $φ(n)\,=\,2\,k\,λ(n)$ for some integer $k\ge1$. Many proofs and statements about RSA, including methods for computing a working $d$ or $e$, can use either $φ$ or $λ$; the results ($d$ or $e$) often differ, but remain correct in the sense of allowing decryption. Using $λ$ yields a condition on $(n,e,d)$ that's necessary, on top of sufficient. The original article uses $φ$. Modern standards tend to use $λ$ (PKCS#1 allows it, FIPS 186-4 requires it).
If $m$ equals $p$ or $q$, then $m$ and $n$ are not co-primes, in this case, does that means decryption will be wrong ?
No, unless $n$ is divisible by the square of a prime (that is when $p=q$ if $n$ is the product of primes $p$ and $q$, as is often considered in RSA). See one of these two questions.
Addition following comment
where does $m^{φ(n)}\bmod n=1$ come from?
Consider a finite group with group law $\cdot$ (multiplicative notation). For any of it's element $m$, we can define the order $\operatorname{ord}(m)$ of element $m$ as the smallest strictly positive integer such that $\underbrace{m\cdot m\cdot\ldots\cdot m}_{\operatorname{ord}(m)\text{ terms}}$ is the neutral of the group. By a fundamental theorem due to Lagrange, the order $\operatorname{ord}(m)$ of any element $m$ in a finite group divides the number of elements in the finite group.
The integers $m\in[0,n)$ with $\gcd(m,n)=1$ form a finite group under multiplication modulo $n$ (that's the multiplicative subgroup $\mathbb Z_n^*$ of the finite ring $\mathbb Z_n$, which neutral is $1$). $φ(n)$ is, by definition, the number of such integers $m$, thus the number of elements in that finite group. Thus for all $m$ in that group, it's defined $\operatorname{ord}(m)$ such that $m^{\operatorname{ord}(m)}\bmod n=1$, and exist some integer $\ell$ (dependent on $m$) with $φ(n)=\operatorname{ord}(m)\,\ell$. It follows
$$\begin{align}
m^{φ(n)}\bmod n&=m^{\operatorname{ord}(m)\,\ell}\bmod n\\
&=\left(m^{\operatorname{ord}(m)}\right)^\ell\bmod n\\
&=1^\ell\bmod n\\\
&=1
\end{align}$$
