Does mean=mode imply a symmetric distribution? I know this question has been asked with the case mean=median, but I did not find anything related to mean=mode.
If the mode equals the mean, can I always conclude this is a symmetric distribution? Will I be forced to know also the median for this manner?
 A: Mean = mode doesn't imply symmetry.
Even if mean = median = mode you still don't necessarily have symmetry. 
And in anticipation of the potential followup -- even if mean=median=mode and the third central moment is zero (so moment-skewness is 0), you still don't necessarily have symmetry.
... but there was a followup to that one. NickT asked in comments if having all odd moments zero was enough to require symmetry. The answer to that is also no. [See the discussion at the end.$^\dagger$]
Those various things are all implied by symmetry (assuming the relevant moments are finite) but the implication doesn't go the other way - in spite of many an elementary text clearly saying otherwise about one or more of them.
Counterexamples are pretty trivial to construct. 
Consider the following discrete distribution: 
  x     -4    0    1    5
P(X=x)  0.2  0.4  0.3  0.1

It has mean, median, mode and third central moment (and hence moment-skewness) all 0 but it is asymmetric.

This sort of example can be done with a purely continuous distribution as well. For example, here's a density with the same properties:

This is a mixture of symmetric triangular densities (each with range 2) with means at 
-6, -4, -3, -1, 0, 1, 2, 5 and mixture weights 0.08, 0.08, 0.12, 0.08, 0.28, 0.08, 0.08, 0.20 respectively. The fact that I just made this now -- having never seen it before -- suggests how simple these cases are to construct.
[I chose triangular mixture components in order that the mode would be visually unambiguous -- a smoother distribution could have been used.]

Here's an additional discrete example to address Hong Ooi's questions about how far from symmetry these conditions allow you to get. This is by no means a limiting case, it's just illustrating that it's simple to make a less symmetric looking example:
   x    -2    0    1    6
P(X=x) 0.175 0.5  0.32 0.005


The spike at 0 can be made relatively higher or lower without changing the conditions; similarly the point out to the right can be placed further away (with a reduction in probability) without changing the relative heights at 1 and -2 by much (i.e. their relative probability will stay close to the 2:1 ratio as you move the rightmost element about).

More detail on the response to NickT's question
$\dagger$ The all-odd-moments zero case is addressed in a number of questions on site. There's an example here (see the plot) based on the details here (see toward the end of the answer). That is a continuous unimodal asymmetric density with all odd moments 0 and mean=median=mode. The median is 0 by the 50-50 mixture construction, the mode is 0 by inspection -- all members of the family on the real half-line from which the example is constructed have a density that's monotonic decreasing from a finite value at the origin, and the mean is zero because all odd moments are 0.
A: To repeat an answer I gave elsewhere, but fits here too:
$$\mathbb{P}(X=n) = \left\{
     \begin{array}{ll}
       0.03 &  n=-3 \\
       0.04 &  n=-2 \\
       0.25 &  n=-1 \\
       0.40 &  n=0 \\
       0.15 &  n=1 \\
       0.12 &  n=2 \\
       0.01 &  n=3  
     \end{array}
   \right.$$

which not only has mean, median and mode all equal, but also has zero skewness.  Many other versions are possible.
A: Try this set of numbers:
\begin{align}
X &= \{2,3,5,5,10\}  \\[10pt]
{\rm mean}(X) &= 5  \\
{\rm median}(X) &= 5  \\
{\rm mode}(X) &= 5
\end{align}

I wouldn't call that distribution symmetrical.
A: No.
Let $X$ be a discrete random variable with $p(X = -2) = \tfrac{1}{6}$, $p(X = 0) = \tfrac{1}{2}$, and $p(X = 1) = \tfrac{1}{3}$. Obviously, $X$ is not symmetric, but its mean and mode are both 0.
